WBCHSE Class 12 Physics Electric Energy And Power Notes

Electric Energy And Power Introduction

In the preceding chapters, different types of electrical circuits, their constructions and uses have been discussed. The subject of discussion of this chapter is the measurement of electrical energy consumed in different electrical circuits and the power of different effective elements of a circuit. The practical idea about electrical power and energy is essential for systematic planning and control of the following subjects

1. How much electrical energy is to be generated in a generating station,
2. Howdie energy is to be distributed in different areas,
3. How the loss of energy is to be minimized,
4. What types of electrical appliances are to be used in the residence office and factory and how they should be used so that, necessary work will be obtained at minimum cost?

Electric Energy And Power Electrical Work

From the discussion in statical electricity, we know that, if Q amount of charge flows through a section under the potential difference V, then the amount of electrical work done is given by,

W = Q x V….(1)

In this equation, if Q is measured in coulomb'(C) and V in volt (V), then the unit of IV is the joule (I), i.e., the relation of the unit is

I = C x V

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Definition: The amount of work done to carry a 1-coulomb charge through a potential difference of 1 volt is called 1 joule.

Besides this, we know that, if current I flow through a conductor for time t, then the amount of charge flowing through it is given by,

Q = It…..(2)

So, combining equations (1) and (2) wo haw.

The electrical work done in the conductor,

W = VIt…..(3)

This work is the amount of electrical energy consumed by the conductor.

Active and Passive Electrical Devices:

The electrical work done by an electrical source in a circuit, l.e., the electrical energy sent in a circuit is used up generally in two ways:

Electrical energy is converted only to treat energy in the connecting wires and in some devices of the circuit.

In most of the cases, this heat energy has no use, i.e., this energy is dissipated. These devices are called passive devices of the circuit and their resistances are called passive resistances.

But in our daily lives, heat energy is utilized in many ways; e.g., electric heaters, electric irons, etc. are very useful appliances.

Yet, the resistance of diesel appliances will also be called passive resistance in conformity with the present discussion.

Electrical energy may be converted to other forms of energy.

For Example, from electrical energy, mechanical energy is obtained in an electric fan and chemical energy is obtained during the charging of a storage cell.

These types of devices are called active devices. However, in these appliances, some amount of heat is evolved. So, it may be said that the active devices possess some passive resistance also.

Let V be the potential difference between the points ll and C in the circuit. If current 1 flows through the circuit for time t, then electrical work,

IV = Vlt

In section AB of the circuit, there is an active electrical device (e.g., an electric fan) and in tire section DC there is a passive device whose resistance is R . This R also Includes (he resistance that generates heat in the active device.

Now, V_A – V_B = V₀

and, V_B – V_C = V’

V_AC = V_A – V_C

= (V_A– V_B) + (V_B– V_C)

= V₀ + V’

= V (say)

∴ V = V₀ + V’

or, Vlt = V₀lt+ V’lt

Again, in the section BC, according to Ohm’s law,

V’ = IR

or, V’t = IR- It

= I²Rt

So, Vlt= V₀It+I²Rt….(4)

This equation (4) can be expressed in the following form:

Electrical work in the circuit = transformed active energy + exothermic energy

If energy and heat are expressed in the same unit (e.g., in SI, both have the same unit joule or), the amount of exothermic energy and heat produced (H) will be equal

i.e., in that case, H = I²Rt

But in many cases, calorie (CGS) is used as the unit of heat. In that case, Joule’s law regarding the relation between heat and work, i.e., W = JH is to be used. In that case,

J = mechanical equivalent of heat or Joule’s equivalent
of heat

= 4.2 J.cal^-1

= 4.2 x 10⁷ erg.cal^-1

In this condition, if exothermic work or energy is W’, then heat produced,

⇒ \(H=\frac{W^{\prime}}{J}\)

i.e., \(H=\frac{I^2 R t}{J}\)….(5)

If work or energy is expressed in joule (J) and heat in calorie (cal), then J = 4.2 J.cal^-1

i.e., \(\frac{1}{J}=\frac{1}{4.2}\)

So, the practical form of the equation (5) is

⇒ \(H=\frac{t^2 R t}{4.2}=0.24 I^2 \mathrm{Rt}\)….(6)

the expense of electrical work Is known as the soulful effect or Joule heating.

Three laws regarding Joule effect are offlined easily from equation (5) or (6):

⇒ \(H \propto I^2\); when R and t are constants;

⇒ \(H \propto R\); when I and t are constants;

⇒ \(H \propto t\); when I and R are constants.

Mechanical equivalent of heat:

Prom equation (5), we have,

⇒ \(J=\frac{I^2 R t}{H}\)

Now, if I = 1 , R = 1 and t = 1

we have, J = \(\frac{1}{H}\)

From the tills relation, the mechanical equivalent of heat Is defined in current electricity.

Definition: The reciprocal of the heat produced in one second in a conductor of unit resistance, for the passage of unit current through it, is called the mechanical equivalent of heat.

In electricity, J = 4.2 J.cal^-1 means:

The amount of heat produced in 1Ω resistance, for the passage of 1 A current for 1 s through It = \(\frac{1}{4.2}\) cal = 0.24 cal.

Electric Energy And Power Numerical Examples

Example 1. 2A current was sent through a coil of resistance 100Ω for 30 min. Determine the amount of heat produced, the quantity of charge passed, and the amount of work done.
Solution:

Here, I = 2A, R = 100Ω t = 30min = 30 x 60s

∴ The amount of heat produced,

⇒ \(H=\frac{I^2 R t}{J}=\frac{(2)^2 \times 100 \times 30 \times 60}{4.2}\)

= 1.71 x 10⁵ cal

The quantity of charge flowing,

Q = It

=2 x 30 x 60

= 3600C

The amount of work done,

W = QV = QIR

= 3600 x 2 x 100

= 7.2 x 10s J

Example 2. Two separate circuits are made with resistances r₁ and r₂ connected to the same storage battery. What should be the internal resistance (r) of the storage battery for which an equal amount of heat is produced in the external circuits?
Solution:

For the first connection, current Ix = \(I_1=\frac{E}{r+r_1}\) [E = emf of the battery]

So, the heat produced in resistance rx in time t,

⇒ \(H_1=\frac{I_1^2 r_1 t}{J}=\frac{E^2 r_1 t}{\left(r+r_1\right)^2 J}\)

Similarly, the heat produced at the same time in the resistance r₂,

⇒ \(H_2=\frac{E^2 r_2 t}{\left(r+r_2\right)^2 J}\)

According to the question, H₁ = H₂

∴ \(\frac{E^2 r_1 t}{\left(r+r_1\right)^2 J}=\frac{E^2 r_2 t}{\left(r+r_2\right)^2 J}\)

or, \(r_1\left(r+r_2\right)^2=r_2\left(r+r_1\right)^2\)

or, \(r_1\left(r^2+2 r r_2+r_2^2\right)=r_2\left(r^2+2 r r_1+r_1^2\right)\)

or, \(r_1 r^2+r_1 r_2^2=r_2 r^2+r_2 r_1^2\)

or, \(r^2\left(r_1-r_2\right)=r_1 r_2\left(r_1-r_2\right)\)

or, \(r^2=r_1 r_2\)

or, \(r=\sqrt{r_1 r_2}\)

Example 3. A heating coll of resistance 5Ω is connected to a cell. The internal resistance of the cell is 20Ω. Calculate the value of the shunt to be introduced, so that, the energy consumed in the heating coil will be \(\frac{1}{9}\)th of the previous value.
Solution:

Let E be the emf of the cell and S be the shunt.

Here, the internal resistance of the cell, r = 20Ω, and external resistance, R = 5Ω.

In the absence of the shunt, current flowing in the circuit,

⇒ \(I_1=\frac{E}{R+r}\)

In time t, energy consumed in the resistance R,

⇒ \(W_1=I_1^2 R t=\frac{E^2 R t}{(R+r)^2}\)

Now, if shunt S is connected, current flowing in the circuit,

⇒ \(I_2=\frac{E}{\frac{R S}{R+S}+r}=\frac{E(R+S)}{R S+r(R+S)}\)

∴ Current flowing in the resistance R,

⇒ \(I_R=I_2 \cdot \frac{S}{R+S}=\frac{E S}{R S+r(R+S)}\)

∴ In time t, energy consumed in the resistance R

⇒ \(W_2=I_R^2 R t=\frac{E^2 S^2 R t}{[R S+r(R+S)]^2}\)

According to the question,

⇒ \(W_2=\frac{W_1}{9} \quad \text { or, } W_1=9 W_2\)

∴ \(\frac{E^2 R t}{(R+r)^2}=9 \cdot \frac{E^2 S^2 R t}{[R S+r(R+S)]^2}\)

or, \(\frac{1}{(R+r)^2}=\frac{9 S^2}{[R S+r(R+S)]^2}\)

or, \(\frac{1}{R+r}=\frac{3 S}{R S+r(R+S)} \text { or, } R+r=\frac{R S+r R+r S}{3 S}\)

or, \(S=\frac{r R}{2(R+r)}=\frac{20 \times 5}{2(20+5)}=\frac{20 \times 5}{2 \times 25}=2 \Omega\)

Example 4. The rate of energy consumed in 5Ω resistance is 10 J.s^-1. What will be the rate of energy consumed in 4Ω resistance?
Solution:

In the adjoining if, V_A-V_B= V, then,

⇒ \(I_1=\frac{V}{5} \text { and } I_2=\frac{V}{4+6}=\frac{V}{10}\)

∴ \({Energy consumed in 1 \mathrm{~s} in 4 \Omega resistance}{Energy consumed in 1 \mathrm{~s} in 5 \Omega resistance}\)

⇒ \(\frac{I_2^2 \cdot 4}{I_1^2 \cdot 5}=\frac{\left(\frac{V}{10}\right)^2 \cdot 4}{\left(\frac{V}{5}\right)^2 \cdot 5}=\frac{1}{5}\)

So, energy consumed in 1s in 4 resistance

= \(\frac{1}{5}\) x energy consumed in 5Ω resistance

= \(\frac{1}{5}\) x 10

= 2J

i.e., the rate of energy consumed in 4Ω resistance is 2 J.s-1

Example 5. Water boils in an electric kettle for 10 minutes after being switched on. How will you modify the heating coil to boil water in 5 minutes using the same source
Solution:

For the source of power of the same emf, V = constant.

As an equal amount of heat is produced in the two cases, we have

⇒ \(\frac{V^2 t_1}{R_1}=\frac{V^2 t_2}{R_2} \text {, because heat produced } \propto \frac{V^2 t}{R}\)

∴ \(R_2=R_1 \frac{t_2}{t_1}=R_1 \times \frac{5 \mathrm{~min}}{10 \mathrm{~min}}=\frac{1}{2} R_1\)

This means that a heating coil with half the resistance compared to the initial one, is to be used.

Electric Energy And Power Electrical Power

If an electrical source sends current I for time t, in a circuit under potential difference V, then the electrical work done in the circuit,

W = Vlt

We know that the rate of work done with respect to time is called power.

∴ \(P=\frac{W}{t}=\frac{V I t}{t}\)

or, P = VI …(1)

i.e., electrical power = potential difference x current

Unit of power: The units of current and potential difference are volt (V) and ampere (A) respectively. Again, the unit of work is joule (J) and the unit of power is joule/second (J/s) or watt (W).

So, from equation (1) we have,

watt = volt x ampere ….(2)

Definition: If the potential difference at the two terminals of an electrical appliance is 1 volt and the current passing through it is 1 ampere, then the power of the appliance is 1 watt.

We have seen that the relation between the electrical work done in an electrical circuit or in a portion of the circuit is electrical work = active energy + exothermic energy

or, Vlt = V₀It + I2Rt

For t= 1s; VI = V₀I+I2R

Equation (3) expresses that, electrical power applied = active power + power consumed in the production of heat

The active energy/power is of little concern to us. We are only interested in the latter, for which we shall find an alternative expression.

An alternative expression of the power of passive resistance: If V’ is the terminal potential difference and I is the current flowing through a passive resistance R, then according to Ohm’s law,

V’ = IR

The power consumed in this resistance is,

P’ = I²R [equation (3)]

= IR.I

= V’I, as already shown.

Since, \(I=\frac{V^{\prime}}{R}\)

∴ \(P^{\prime}=\left(V^{\prime} \cdot \frac{V^{\prime}}{R}\right)=\frac{V^{\prime 2}}{R}\)

So, \(P^{\prime}=V^{\prime} I=I^2 R=\frac{V^{\prime 2}}{R}\) …(4)

In most electrical circuits, there is no active device. All the resistances are passive. So, the entire power applied is consumed in the form of heat. In that case, the applied potential difference and the potential difference across the passive resistance (or resistances) become equal to each other.

That is, in this condition, V = V’ and P = P’. So equation (4) can be written as,

⇒ \(P=V I=I^2 R=\frac{V^2}{R}\)…(5)

It is evident from equations (4) and (5) that, the power consumed in the external resistance can be expressed in three different ways. P = VI is the general expression of power and it is applicable in all types of electrical appliances.

On the other hand, the relations \(P=I^2 R \text { or, } P=\frac{V^2}{R}\) are used as the expressions for power consumed in exothermic devices.

Wastage of energy in the transmission of electric power. Suppose, electrical power is to be transmitted over long distances from a generating station. Usually, a long line wire is used for this purpose.

Now, as current flows through this wire, a huge amount of heat is generated due to the Joule effect, i.e., a large amount of electrical energy is wasted as heat during transmission. So, the most important condition for transmitting the electrical power from one place to another is that the wastage of energy should be reduced as much as possible.

A solution to minimize wastage of energy: From the expression of the power P = VI, it can be easily understood that the higher the value of V, the lower will be the value of I.

Again, from equation (3), it is seen that, if the value of I is small, the value of power consumed in the production of heat (I2R) will also diminish, i.e., the wastage of energy will also diminish. So, a high-voltage power transmission allows for lesser resistive losses over long distances in wiring.

In actual practice, electrical energy is transmitted along the long line wire at a high voltage, 11000 V or more. At the place where the supply is required, the high voltage is stepped down to 220 V or 440 V by a step-down transformer.

Horsepower or hp: This is a large unit of power. From the discussion of mechanical energy and power, we know,

1 hp = 746 W

Generally, a horsepower unit is used to express the power of those machines (motor, pump, etc.), where electrical energy is transformed into mechanical energy.

Power Consumed in an Electrical Circuit:

Suppose, in an electrical circuit, the emf of the battery is E. If the internal resistance of the battery is r and R is the equivalent resistance of all the resistances used in the external circuit, current in the circuit, \(I=\frac{E}{R+r}\)

∴ Power consumed in the circuit, \(P_0=I^2(R+r)=\frac{E^2}{R+r}\)

Maximum power in the external circuit: A part of the total power, in the circuit equivalent to I²r, is consumed due to the internal resistance of the battery and it cannot be utilized in any other way. The best part of the power is spent in the external circuit and is utilized for running different electrical appliances connected in the external circuit. Since the equivalent resistance of the external circuit is R, the available power in the external circuit,

⇒ \(P=I^2 R=\frac{E^2 R}{(R+r)^2}\)

In case of constant emf and internal resistance of the battery (E = constant, r = constant), the condition of availability of maximum power in the external circuit is given by,

⇒ \(\frac{d P}{d R}=0 \quad \text { or, } \frac{d}{d R}\left[\frac{E^2 R}{(R+r)^2}\right]=0\)

or, \(E^2 \cdot \frac{1(R+r)^2-2(R+r) R}{(R+r)^2}=0\)

or, (R + r)²-2R²-2Rr = 0

or, r²-R² = 0

or, R = r

i.e., maximum power is obtained in the external circuit, if the equivalent resistance of the external circuit is equal to the internal resistance of the battery. The value of maximum power is

⇒ \(P_{\max }=\frac{E^2 R}{(R+R)^2}=\frac{E^2}{4 R}=\frac{E^2}{4 r}\)

Electric Energy And Power Numerical Examples

Example 1. A 20Ω resistor can dissipate a maximum of 2kW power as heat without being damaged. Should this resistor be connected directly across a 300 V dc source of negligible Internal resistance?
Solution:

If the resistor is connected directly with a 300Vdc source, its power would be,

⇒ \(P=\frac{V^2}{R}=\frac{(300)^2}{20}\)

= 4500W

= 4.5W

Since, this power is greater than 2 kW, so the resistor should not be directly connected to a 300V dc source.

Example 2. Three resistors of equal resistances which en connected in series across voltage sources, dissipate 1.00 watatt of power. What would be the power dissipated, if the resistors are connected in parallel across the same source of emf
Solution:

If each resistance is r, then the equivalent resistance in series, R₁ = 3r; and the equivalent resistance in parallel

⇒ \(R_2=\frac{r}{3}\)

In both combinations, the potential difference across the two ends is equal to V (say).

∴ In case of series combination, power, \(P_1=\frac{V^2}{R_1}\) and in case of parallel combination, power, \(P_2=\frac{V^2}{R_2}\)

∴ \(\frac{P_1}{P_2}=\frac{R_2}{R_1} \quad \text { or, } P_2=P_1 \cdot \frac{R_1}{R_2}=100 \times \frac{3 r}{\frac{r}{3}}\)

= 900W

Example 3. The coil of a heater connected to a 200V line consumes a power of 100 W. The coil Is divided Into two equal parts. The two parts are combined In parallel and connected to a 200 V line. What will be the power consumed by the new combination?
Solution:

If R is the resistance of the coll, then power summed in the potential difference V, \(P=\frac{V^2}{R}\)

In the first case, \(100=\frac{(200)^2}{R}\)

In the second case, the resistance of each of the two equal parts = \(\frac{R}{2}\)

So, the equivalent resistance in the parallel combination

⇒ \(\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}=\frac{R}{4}\)

So, power consumed,

⇒ \(P_2=\frac{(200)^2}{\frac{R}{4}}=4 \times \frac{(200)^2}{R}\)

= 4 x 100

= 400W

Example 4. The power consumed in the circuit. Is 150 W. What is the value of it?

Solution:

Equivalent resistance,

⇒ \(r=\frac{R \cdot 2}{R+2}=\frac{2 R}{R+2}\)

So, power, \(P=\frac{V^2}{r}=\frac{V^2(R+2)}{2 R}\)

or, \(\frac{R+2}{R}=\frac{2 P}{V^2} \quad\)

or, \(1+\frac{2}{R}=\frac{2 \times 150}{(15)^2}=\frac{4}{3}\)

or, \(\frac{2}{R}=\frac{4}{3}-1=\frac{1}{3}\)

or, R = 6Ω

Example 5. With a ceil of emf 1.5 V and of internal resistance of 0.1Ω when connected with a resistor and an ammeter of negligible resistance in series, the ammeter shows a 2.0 A steady current. Find

1. The rate of energy dissipated within the cell
2. The power consumed in

Solution:

1. The rate of energy dissipated within the cell

= El = 1.5 x 2.0

= 3W

2. \(I=\frac{E}{R+r}\); R = resistance of the resistor

or, \(R=\frac{E}{I}-r\)

= \(\frac{1.5}{2}-0.1\)

= 0.65Ω

∴ Power consumed in the resistor,

= (2.0)² x 0.65

= 2.6 W

Example 6. A balanced Wheatstone bridge has resistances 100Ω, ion, 500Ω, and 50Ω respectively in its four arms. Determine the ratio of powers consumed in its different arms.
Solution:

Wheatstone Bridge has been shown.

Let \(I_1=\frac{V}{P+Q} \text { and } I_2=\frac{V}{R+S}\)

∴ \(\frac{I_1}{I_2}=\frac{R+S}{P+Q}=\frac{500+50}{100+10}=5 \quad \text { or, } I_1=5 I_2\)

∴ The required ratio

⇒ \(I_1^2 P: I_1^2 Q: I_2^2 R: I_2^2 S\)

⇒ \(\left(5 I_2\right)^2 \times 100:\left(5 I_2\right)^2 \times 10: I_2^2 \times 500: I_2^2 \times 50\)

= 2500: 250: 500: 50

=50: 5: 10: 1

Example 7. A factory requires a power of 90kW. The energy is transmitted to the factory through a 2.5 n line wire. If 10% of the power generated is lost in transmission, calculate

1. The transmission line current
2. The potential difference at the power generating station
3. The potential drop due to line resistance

Solution:

Here, \(\begin{aligned}
\frac{\text { power of the generating station }}{\text { power required for the factory }} & =\frac{100}{(100-10)} \\
& =\frac{100}{90}
\end{aligned}\)

or, power of the generating station = 90 x \(\frac{100}{90}\) = 100 kW

Power lost =100-90

= 10kW

= 10000 W

This power is lost in the transmission line.

So, according to the equation, P = I²R

Current in the transmission line,

⇒ \(I=\sqrt{\frac{P}{R}}=\sqrt{\frac{10000}{2.5}}=\sqrt{4000}\)

= 63.25A(approx.)

According to the equation, P = VI,

potential difference at the generating station,

⇒ \(V=\frac{P}{I}=\frac{10000}{63.25}=1581 \mathrm{~V}(\text { approx })\)

Potential drop due to line resistance,

V’ =IR

= 63.25 x 2.5

= 158.1V (approx.)

Example 8. Electrical energy is transmitted at the rate of 2.2 MW through the line wire. The resistance of the line wire is 25Ω. Calculate the percentage of heat dissipation of the electrical energy for each line voltage

1. 22000V
2. 110kV.

Solution:

VI = 2.2 MW

= 2.2 x 106W

1. \(V=22000 \mathrm{~V} ; I=\frac{V I}{V}=\frac{2.2 \times 10^6}{22000}=100 \mathrm{~A}\)

∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I} \times 100 \%\)

⇒ \(\frac{(100)^2 \times 25}{2.2 \times 10^6} \times 100 \%\)

= 11.4%

2. \(V=110 \mathrm{kV}=110000 \mathrm{~V} ; I=\frac{V I}{V}=\frac{2.2 \times 10^6}{110000}=20 \mathrm{~A}\)

∴ Percentage of heat dissipation = \(\frac{I^2 R}{V I} \times 100 \%\)

⇒ \(\frac{(20)^2 \times 25}{2.2 \times 10^6} \times 100 \%\)

0.45%

[Obviously, it is seen that the dissipation of energy becomes less if the line voltage is high.]

Electric Energy And Power Commercial Units Of Electrical Work And Electrical Energy

Watt. hour (W.h): Work done or electrical energy consumed in 1 hour by an electrical appliance having power 1W is called 1W.h.

∴ lW h = lWx lh = 1W x 3600 s =3600J

It is obvious that, W.h = W x h

= V x A x h

i.e., if 1 A current flows for lh under the potential difference of IV, the amount of electrical energy spent will be 1W.h.

kiloWatt. hour (kW.h): Work done or electrical energy spent in lh by an electrical appliance having power lkW is called lkW h or 1 BOT unit.

Obviously, this is a bigger unit of electrical energy. The electricity supplied to us for our use is measured by this unit. Many times, kW h or BOT unit is simply called unit

lkW.h = 1000 W h = 1000 W x lh

= 1000 W X 3600 s

= 3.6 X 10⁶ W.S

= 3.6 X 10⁶J

⇒ \(\mathrm{kW} \cdot \mathrm{h}=\frac{\mathbf{W} \times \mathbf{h}}{1000}=\frac{\mathrm{V} \times \mathrm{A} \times \mathrm{h}}{1000}\)

For Example, if an electric bulb marked 100 W glows for 20h then, the amount of electrical energy spent = \(=\frac{100 \times 20}{1000}=2 \mathrm{~kW} \cdot \mathrm{h}\)

= 2 unit

So, in this case, the value of the electric meter will rise up by 2 units.

Electric Energy And Power Numerical Examples

Example 1. In a house, there are 10 lamps of 40 W each, 5 fans of 80 W each, and a TV set of 80 W. They run for 6 hours a day. Find the consumption of electrical energy in a month of 30 days. What is its value in BOT unit
Solution:

Total power = (10 x 40) + (5 x 80) + 80

= 880 W.

Total time = 6 x 30

= 180 h

= 180 x 3600 s

Energy consumed in a month = 880 x 180 x 360

= 5.7 x 108J

Its value in BOT unit = \(\frac{\mathrm{W} \times \mathrm{h}}{1000}=\frac{880 \times 180}{1000}\)

= 158.4

Example 2. In an evening college, there are 100 bulbs of 60 W each, 80 bulbs of 100 W each, and 70 fans of 100 W each. They run for 5h, 4h and 4h respectively per day. If each kW.h costs ₹ 0.50, calculate the electric bill of the college for a month
Solution:

The electrical energy consumed by 60 W bulb

= \(\frac{(60 \times 100) \times 5 \times 30}{1000}\)

= 900kW.h

Electrical energy consumed by 100 W bulbs

⇒ \(\frac{(100 \times 80) \times 4 \times 30}{1000}\)

= 960kW.h

Electrical energy consumed by fans

⇒ \(\frac{(100 \times 70) \times 4 \times 30}{1000}\)

= 840 kW.h

∴ Cost of electrical energy = (900 + 960 + 840) x 0.50

= ₹ 1350

Example 3. In a house, there are 20 lamps of 60W each, and 10 fans that operate in 0.5A current. If the main power supply is 220V, the expense per kW h is 50 paise and each application runs 6 h per day then calculate the electric bill of
Solution:

November month has 30 days.

∴Total time = 30 x 6h = 180h ;

50 paise = ₹ 0.5

Power of each fan =220 x 0.5 = 110W

∴ Electrical energy

⇒ \(\frac{(60 \times 20+110 \times 10) \times(30 \times 6)}{1000}\)

⇒ \(\frac{2300 \times 30 \times 6}{1000}\)

= 23 x 18 kW.h

∴Electric bill of November = ₹(23 x 18) x 0.5

= ₹ 207

Example 4. There are six 40 W and two 100 W lamps, four 40 W fans, and a 1000 W electric heater in the house. If in April, each lamp runs for 5 hours a day, each fan for 15 hours a day, and the heater for 2 hours a day, what will be the electric bill for that month? It may be supposed that the main supply voltage is 200 V and the cost of each BOT unit = ₹ 1.50. Which one of the given three fuse wires of rating 5 A, 10 A, and 15 A will be appropriate for connection in the main switch?
Solution:

Number of days in April = 30

Electrical energy consumed by the lamp

⇒ \(\frac{(6 \times 40+2 \times 100) \times 5 \times 30}{1000}\)

= 66 BOT unit

Electrical energy consumed by the fans

⇒ \(\frac{(4 \times 40) \times 15 \times 30}{1000}\)

= 72 BOT unit

The electrical energy consumed by the heat

⇒ \(=\frac{1000 \times 2 \times 30}{1000}=60 \mathrm{BOT} \text { unit }\)

∴ Monthly expenditure on electricity

= (66 + 72 + 60) X 1.50 = ₹ 297

Again, the total power of the appliances

= (6 x 40) + (2 x 100) + (4 x 40) + 1000

= 1600 W

If all the appliances run simultaneously, according to the relation, P = V x I,

we have, \(I=\frac{P}{V}=\frac{1600}{200}=8 \mathrm{~A}\)

So, a 10 A fuse will be appropriate for connection in the main switch
because a 5 A fuse will melt and there is no need for a 15 A fuse

Example 5. The power of a small electric motor is \(\frac{1}{8}\) HP. If it is connected to a 220 V supply line, how much current will it j draw? If the motor runs for 80 hours, what will be the cost? j (Each BOT unit costs 70 paise)
Solution:

Power, P = \(\frac{1}{8}\) HP = \(\frac{1}{8}\) x 746 W

According to the relation, P = V x I,

we have, \(I=\frac{P}{V}=\frac{1}{8} \times 746 \times \frac{1}{220}=0.424 \mathrm{~A}\)

Number of BOT units = \(\frac{1}{8} \times 746 \times 80 \times \frac{1}{1000}\)

∴ \(\text { Cost }=\frac{1}{8} \times 746 \times 80 \times \frac{1}{1000} \times 70 \text { paise }\)

= ₹ 5.22 (approx.)

Example 6. A heating coil of resistance 100 fl is connected for 30 min to 220 V. By this time, determine

1. Amount of charge flowing,
2. Amount of electrical energy consumed
3. Amountofheatgenerated. Determine the cost of consumed electrical energy if 1 kW.h costs ₹1.

Solution:

1. Amount of charge flowing,

⇒ \(Q=I t=\frac{V}{R} t=\frac{220}{100} \times 30 \times 60=3960 \mathrm{C}\)

2. Amount of electrical energy consumed,

W = Q.V

= 3960 x 220

= 8.712 x 105 J

3. Amount of heat generated,

⇒ \(H=\frac{W}{J}=\frac{8.712 \times 10^5}{4.2}=2.074 \times 10^5 \mathrm{cal}\)

= 2.074 x 105 cal

4. Cost of consumed electrical energy = \(\frac{8.712 \times 10^5}{3600 \times 1000} \times 1\)

= ₹ 0.24

= 24 paise

Electric Fuse:

Every metal, such as copper, aluminum, etc. used as the element of connecting wires in electrical circuits and electrical applications has a definite melting point.

Due to a short circuit or any other reason, if an excessively high current flows in the circuit, the temperature of the wire may rise to its melting point.

As a result, the whole circuit may be damaged and may cause fire. A fuse inserted in the circuit may prevent an accidental fire. It is prepared from an alloy (e.g., 3 parts lead and 1 part tin) having a comparatively low melting point.

This fuse is placed in an insulating box and is connected in series with the main circuit. Before the current of the main circuit reaches the danger point, the temperature of the fuse reaches its melting point.

As a result, the fuse melts down and the main circuit is disconnected. Hence, there is no chance of damage to the main circuit

Circuit breaker:

This is an automatic switch system that disconnects the circuit before being damaged due to short-circuit or overload. At the very moment the circuit is disconnected, the switch of the breaker moves to the ‘off’ position.

After replacing the defective instrument for which the circuit was broken, the switch is made ‘on current flows as usual through the circuit. In different types of circuit breakers, different physical phenomena are used.

For Example, in the circuit through which high current flows, a bimetallic strip bends due to heating and disconnects the circuit; or if a high current flows through an electromagnet, it attracts a soft-iron plate of the circuit and thus the circuit is cut off.

The main difference between a fuse wire and a circuit breaker is that, the fuse wire is burnt off and needs to be replaced with a new wire every time the fuse melts.

On the other hand, a circuit breaker is not at all damaged and hence there is no need to replace it. Once the fault has been repaired, the contacts must again be closed to restore power to the interrupted circuit.

The circuit breaker which is used in low voltage lines (0-1000 V) is generally called a Mechanical or Miniature Circuit Breaker (MCB).

Nowadays, in place of fuse wire, MCB is widely used. The switch gear is generally used as a circuit breaker in the supply line of high voltage (11000 V or more) from the generating station.

Highest safe current: If l is the length of the fuse wire, r is its radius, and \(\rho\) its specific resistance, then the resistance of the wire is given by,

⇒ \(R=\rho \frac{l}{\pi r^2}\)

So, heat generated in the wire per second is,

⇒ \(H=\frac{l^2 R}{J}=\frac{l^2 \rho l}{J \pi r^2}\)

This heat is used in two ways:

1. One part increases the temperature of the tire wire,

2. The other part radiates to the surroundings from the outer surface of the wire.

Thus, a situation arises when the total heat generated in the wire, radiates to the surroundings. So, the temperature does not increase further and reaches a fixed value. If this temperature lies just below the tire melting point of the material of the fuse wire, then the tire current flowing in the tire wire at this stage is called the tire’s highest safe current.

Now, the outer surface area of the tire wire = 2πrl.

If h is the tire amount of radiant heat from the unit area of the tire wire per second, then the total radiant heat from the outer surface of the wire per second is, H’ = 2πrlh.

So, for the highest safe current I,

⇒ \(H=H^{\prime}\)

or, \(\frac{I^2 \rho l}{J \pi r^2}=2 \pi r l h \quad\)

or, \(I^2=\frac{2 \pi^2 J h}{\rho} \cdot r^3\)

i.e., \(I^2 \propto r^3 \text { or, } I \propto r^{3 / 2}\)

It is to be noted that, the expression for highest safe current I is independent of the length I of the fuse wire.

Electric Energy And Power Rating Of Electrical Instruments

Voltage rating: Bach and every electrical instrument depends on potential differences. Each instrument is marked with a fixed value of potential difference, indicating the range of potential difference within which it can be run safely. This is called voltage rating.

For Example, the voltage rating of each domestic appliance is 220 V. This means that, if the potential difference between two ends of the appliance is 220 V, then its efficiency becomes maximum.

The efficiency of the appliance decreases if for any reason the supply voltage Is diminished, i.e., the electric lamp glows dimly, the electric fan moves comparatively slowly, etc.

On the other hand, if the voltage exceeds 220 V, there is a chance of damage to the appliance.

This is to note that, the appliances heaving the same voltage rating should be connected in parallel while used simultaneously in a circuit. In that case, the potential difference across each application remains the same.

For this reason, daily appliances are all connected in parallel combination with a 220 V supply line.

Watt rating: Different electrical appliances, even running at the same voltage may not consume the same electrical energy in unit time.

Hence, the power of the appliance, i.e., the electrical energy consumed per second should be mentioned.

This is called the watt rating of electrical appliances. Thus 220V-100W on the body of a lamp or 220V-2000W on the body of a heater refers to the voltage and watt ratings of the appliance.

Significance of ratings:

1. Maximum capability: Voltage rating indicates at what voltage, the appliance does maximum work, and watt rating indicates the rate of consumption of electrical energy at that voltage.

For Example, 220V-100W indicates diet, if the voltage at the two ends of the lamp is 220 V, it will glow with maximum brightness, consuming energy at the rate of 100 W

2. Calculation of current and resistance: It is possible to calculate the current and resistance of the appliance when it operates with maximum efficiency.

Significance of rating Example:

⇒ \(I(\text { current })=\frac{P(\text { power })}{V(\text { potential difference })}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}=0.45 \mathrm{~A}\)

⇒ \(R(\text { resistance })=\frac{V}{I}=\frac{220 \mathrm{~V}}{0.45 \mathrm{~A}}=489 \Omega\)

Safety of supply line: Whether an electrical appliance is safe or not for a supply line, is known from its rating. Consider the case of a 220 V-2000 W electric heater.

Current drawn by the heater = \(\frac{2000}{20}\) = 9.1 A (approx.)

So, this heater cannot be used in the house, where the meter and fuse are installed for 5 A current. If it is used, the supply line will be damaged.

Power in a combination of more than one electrical device:

Parallel combination: Suppose, and R₂ are the resistances of two electrical appliances. Their watt ratings are P₁ and P₂ respectively and the voltage rating of both of them is V.

If they are connected in parallel to an electric source of voltage V, then the voltage at the two ends of both the appliances will be V. The two appliances will function at full power and the total power consumed by the circuit will be,

P = P₁ + P₂

For any number of electrical appliances connected in parallel, P = P₁+P₂ + P₃+…. obviously, total power will be greater than that of each appliance.

Household electrical appliances are usually connected in
parallel.

Now, \(P_1=\frac{V^2}{R_1} \text { and } P_2=\frac{V^2}{R_2}\)

∴ \(\frac{P_1}{P_2}=\frac{R_2}{R_1}\)….(1)

For Example, if two electric lamps of 220 V-100 W and 220 V-60 W are connected in parallel and the combination is joined to a 220 V line, the 100 W lamp will glow brighter and the total power of the circuit will be, P – 100 + 60 = 160 W.

Series combination: Now the two appliances are nected in series and this combination is connected to the same electric source. The same current, say I, will flow through both of them. So, their powers will be respectively,

P’1 = I²R₁ and P’2 = I²R₂

So, \(\frac{P_1^{\prime}}{P_2^{\prime}}=\frac{R_1}{R_2}\) …..(2)

From equations (1) and (2) we have,

⇒ \(\frac{P_1^{\prime}}{P_2^{\prime}}=\frac{P_2}{P_1}\)

So, if \(P_1>P_2, P_1^{\prime}<P_2^{\prime}\)

i.e., in series combination, the appliance having a higher watt rating will consume less power.

For Example, if two electric lamps having the power of 100 W and 60 W are connected in series, the power of the 100 W lamp will be comparatively lower and it will glow with less brightness.

Again, according to,

⇒ \(I=\frac{V}{R_1+R_2}\)

So, the total power of the circuit,

⇒ \(P^{\prime}=I^2\left(R_1+R_2\right)=\frac{V^2}{R_1+R_2}\)

∴ \(\frac{1}{P^{\prime}}=\frac{R_1+R_2}{V^2}=\frac{R_1}{V^2}+\frac{R_2}{V^2}=\frac{1}{P_1}+\frac{1}{P_2}\)

For any number of electrical appliances connected in series,

⇒ \(\frac{1}{P^{\prime}}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots\)

Obviously, the total power P’ will be less than that of each application.

For Example, if two electric lamps of 220 V-100 W and 220 V-60 W are connected in series and joined to a 220 V supply line, the 60 W lamp will glow brighter and the total power of the circuit will be less than 60 W. Here

⇒ \(\frac{1}{P^{\prime}}=\frac{1}{100}+\frac{1}{60} \quad\)

or, \(P^{\prime}=\frac{100 \times 60}{100+60}\)

= 37.5 W^

Rating of a resistor: No resistor can endure excess current over a certain designated value depending upon its nature. If the limit is exceeded, the excessive heat generated in the resistor will bum it. We may say that, as soon as the power consumption due to the production of heat crosses a certain limit, the resistor gets damaged. This maximum power tolerance is known as the watt rating of the resistor.

For Example, a resistor with a rating lW-100Ω means a 100Ω resistance and of maximum power tolerance 1W. We know, power,

P = VI = IR.I = I²R

So maximum current tolerance of the resistor Is,

⇒ \(I=\sqrt{\frac{P}{R}}=\sqrt{\frac{1}{100}}=\frac{1}{10} \mathrm{~A}\)

Under this condition, the potential difference between the two ends of the resistor is,

⇒ \(V=I R=\frac{1}{10} \times 100\)

= 10V

Therefore, before joining a circuit, the rating of a resistor should be known. In that case, it can be connected in such a way that the current should always remain below

⇒ \(\frac{1}{10}\) A through it

Electric Energy And Power Numerical Examples

Example 1. A 220V-60W electric bulb is connected To a 220 V line. What is the resistance of the filament of the bulb, when it is turned on?
Solution:

Power, \(P=V I=V \cdot \frac{V}{R}=\frac{V^2}{R} \quad[∵ V=I R]\)

∴ \(R=\frac{V^2}{P}=\frac{220 \times 220}{60}\)

= 806.67Ω

Example 2. The resistance of a hot tungsten filament is about 10 times that in its normal state. What will be the resistance of a 100 W-200 V tungsten lamp in its normal state?
Solution:

Resistance of the lamp when it is hot,

⇒ \(R=\frac{V^2}{P}=\frac{(200)^2}{100}=400 \Omega\)

So, resistance of the lamp in its normal state = \(\frac{400}{10}=40 \Omega\)

Example 3. The main meter of a house is marked 10 A-220 V. How many 60 W electrical amps can be used safely in this line?
Solution:

Maximum power of the household line,

P = VI

= 220 X 10

= 2200 W

Maximum number of 60 W lamp = \(\frac{2200}{6}\) = 36.67

So, a maximum of 36 lamps can be used at a time.

Example 4. A 220V-100W electric lamp fuses above 150 power. What should be the maximum tolerable voltage for the lamp?
Solution:

Power, \(P=\frac{V^2}{R}\)

If we ignore the change or resistance to the Increase of power,

We have, \(P \propto V^2\)

So, for two different powers, \(\frac{P_1}{P_2}=\frac{v_1^2}{v_2^2}\)

i.e., \(\cdot V_2=V_1 \sqrt{\frac{P_2}{P_1}}=220 \times \sqrt{\frac{150}{100}}\)

= 269.4 V

So, the maximum tolerable voltage for the lamp Is 269.4 V.

Example 5. In order to run a 60V-120W lamp safely in a 220V dc line, a resistor of what minimum magnitude should be placed in series with it?
Solution:

When the lamp runs safely In the given dc line, the maximum current is,

⇒ \(I=\frac{120}{60}=2 \mathrm{~A}\)

So, this 2 A current will also flow through the resistance placed in series.

Again, the potential difference at the two ends of the resistance

= 220-60

= 160V.

∴ The minimum value of the resistance = \(\frac{160}{2}\)

= 80Ω

Example 6. Draw a household circuit having a 1200 W toaster, a 1000 W oven, an 800 W heater, and a 1500 W cooler. The circular has a heavy-duty wire and a 20 A circuit breaker. Will the circuit breaker trip, if all the appliances are operated simultaneously in a 220 V supply voltage?
Solution:

The household circuit. When all the appliances are operated simultaneously, the total po is, P = 1200 + 1000 + 800 + 1500

= 4500 W.

Now, P = VI

or, \(I=\frac{P}{V}=\frac{4500}{220}\)

= 20.45A

But the circuit breaker is rated for 20 A.

So, if all the appliances are operated simultaneously, the circuit breaker will trip.

Example 7. Two lamps of 200 W and 100W arc connected in series in 200 V mains. Assuming the resistance of the two lamps remains unchanged, calculate the power consumed by each of them.
Solution:

⇒ \(P=\frac{V^2}{R} \quad \text { or, } R=\frac{V^2}{P}\)

So, the resistance of the first lamp, \(R_1=\frac{200 \times 200}{200}=200 \Omega\)

and resistance of the second lamp, \(R_2=\frac{200 \times 200}{100}=400 \Omega\)

Now, if we connect the two lamps in series with 200 V mains, current flowing through each of them,

⇒ \(I=\frac{V}{R_1+R_2}=\frac{200}{200+400}=\frac{1}{3} \mathrm{~A}\)

∴ Power of the first lamp, \(P_1=I^2 R_1=\frac{1}{3} \times \frac{1}{3} \times 200=22.2 \mathrm{~W}\)

Power of the second lamp, \(P_2=I^2 R_2=\frac{1}{3} \times \frac{1}{3} \times 400=44.4 \mathrm{~W}\)

Example 8. Two electric bulbs each designed to operate with a power of 500 W In a 220 V line are connected in series in a 110 V line. What will be the power generated by each bulb?
Solution:

Power, \(P=\frac{V^2}{R}\)

So, the resistance of each bulb,

⇒ \(R=\frac{V^2}{P}=\frac{(220)^2}{500}=\frac{484}{5} \Omega\)

When connected in series, the potential difference at the two ends of each bulb = \(\frac{110}{2}\)

=55V.

∴ \(\frac{(55)^2}{\frac{484}{5}}=\frac{55 \times 55 \times 5}{484}\)

= 31.25W

Example 9. If the supply voltage drops from 220 V to 200 V , what would be the percentage reduction in heat produced by a 220 V-1000 W heater? Neglect the change of resistance. If the change of resistance is taken into consideration, would the reduction of heat produced be smaller or larger than the previously calculated value?
Solution:

We know, \(P=\frac{V^2}{R}\)

So, \(R=\frac{V^2}{P}\)

If the variation of resistance is ignored,

In the first case, \(R=\frac{V_1^2}{P_1}\)

In the second case, \(R=\frac{V_2^2}{P_2}\)

i.e., \(\frac{V_1^2}{P_1}=\frac{V_2^2}{P_2} \text { or, } \frac{P_2}{P_1}=\frac{V_2^2}{V_1^2}\)

or, \(\frac{P_2-P_1}{P_1}=\frac{V_2^2-V_1^2}{V_1^2}=\frac{\left(V_2+V_1\right)\left(V_2-V_1\right)}{V_1^2}\)

= \(\frac{(200+220)(200-220)}{220 \times 220}\)

or, \(\frac{P_2-P_1}{P_1}=-\frac{420 \times 20}{220 \times 220}\)

= -0.1736 (approx.)

So, the power of the heater will reduce by 17.36%; i.e., the percentage reduction of heat produced is 17.36%. In the above calculation, the decrease of resistance with voltage drop was ignored. But actually, with a decrease in the temperature of the coil, its resistance also decreases. Hence according to the relation P = \(P=\frac{V^2}{R}\), power Increases, i.e., heat supplied by the heater also increases. So, the reduction of heat produced will be lower than 17.36%.

Example 10. The emf of the cell, E = 20V. The rating of each resistance R₁ and R₂, is 1W-100Ω. What should be the minimum value of the resistance R in the circuit? Also, find its minimum watt rating

Solution:

According to the formula,

P = I²R

or, \(I=\sqrt{P / R}\)

The maximum current that can flow through each of the resistance R1, and R2 is,

⇒ \(I_1=I_2=\sqrt{\frac{1}{100}}=\frac{1}{10} \mathrm{~A}\)

∴ The maximum current that can flow through the resistance R,

⇒ \(I=I_1+I_2=2 \times \frac{1}{10}=\frac{1}{5} \mathrm{~A}\)

On die other hand, equivalent resistance of R1 and R2.

⇒ \(\frac{R_1 R_2}{R_1+R_2}=\frac{100 \times 100}{100+100}=50 \Omega\)

Now, \(I=\frac{E}{R+50} \quad\)

or, \(\frac{1}{5}=\frac{20}{R+50}\)

or, R + 50 = 100

∴ R = 100-50 = 50Ω

i.e., to keep the value of I lower than \(\frac{1}{5}\) A, R sliould be greater than 50Ω. Power consumption for the minimum value of R,

⇒ \(P=I^2 R=\left(\frac{1}{5}\right)^2 \times 50=2 \mathrm{~W}\)

Example 11. If a 220 V-1000 W lamp is connected to 110 V line then what will be the power consumed by it?
Solution:

Resistance of the lamp, \(R=\frac{V_1^2}{P}=\frac{(220)^2}{1000} \Omega\)

∴ Power consumed by the lamp when it is connected to 110V
line,

⇒ \(P_2=\frac{V_2^2}{R}=\frac{(110)^2 \times 1000}{(220)^2}=250 \mathrm{~W}\)

Example 12. 2.2 kW power is supplied through a line of 10Ω resistance under a 22000 V voltage difference. What is the rate of heat dissipation in the line?
Solution:

⇒ \(I=\frac{P}{V}=\frac{2.2 \times 10^3}{22000} \quad\left[∵ 2.2 \mathrm{~kW}=2.2 \times 10^3 \mathrm{~W}\right]\)

= 0.1 A

∴ Rate of heat dissipation = I²R

= (0.1 )2 x 10

= 0.1 W

Example 13. The potential difference between the two ends of an electric lamp is decreased by 1%. Neglecting the change in its resistance, calculate the percentage increase or decrease in the power of the lamp.
Solution:

We know power, \(P=\frac{V^2}{R}\)

∴ InR = 21nV-InR

By differentiating \(\frac{d P}{P}=2 \cdot \frac{d V}{V}[∵ R=\text { constant, } d R=0]\)

According to die problem, \(\frac{d V}{V}=-1 \%=-\frac{1}{100}\)

∴ \(\frac{d P}{P}=-\frac{2}{100}=-2 \%\)

i.e., the power of the lamp will be decreased by 2%.

Example 14. 15 kW power is supplied through a line of 0.5Ω resistance under 250 V potential difference. Find the efficiency of the supply in percentage.
Solution:

Power, P = 15 kW

= 15000 W

So, currently in the supply line.

⇒ \(I=\frac{P}{V}=\frac{15000}{250}=60 \mathrm{~A}\)

Power loss due to inactive resistance in the line

= I²R = (60)² x 0.5

= 1800W;

Therefore, total power in line = 15000 + 1800

= 16800 W

∴ \(\text { Efficiency }=\frac{\text { effective power }}{\text { total power }} \times 100 \%\)

⇒ \(\frac{15000}{16800} \times 100 \%=89 \%\)

Example 15. Two incandescent lamps (25 W, 120 V) and (100 W, 120 V) are connected in series across a 240 V supply. Assuming that the resistances of the lamps do not vary with current, find the power dissipated in each lamp after the connection.
Solution:

The resistance of the lamp is 25 W-120 V,

⇒ \(R_1=\frac{120^2}{25}=576 \Omega\)

The resistance of the lamp \(100 \mathrm{~W}-120 \mathrm{~V}, R_2=\frac{120^2}{100}=144 \Omega\)

When the lamp is connected in series, the equivalent resistance of the circuit,

R’ = 576 + 144 = 720Ω

∴ Current flowing through the circuit = \(=\frac{240}{720}=0.33 \mathrm{~A}\)

∴ Power dissipated in die first bulb =(0.33)²x576 = 62.72 W

∴ Power dissipated in die second bulb = (0.33)² x 144

= 15.68 W

Electric Energy And Power Synopsis

• The amount of work done to carry a 1-coulomb charge through a potential difference of 1 volt is called 1 joule.
• The reciprocal of the heat produced in one second in a conductor of unit resistance, for the passage of unit current through it, is called the mechanical equivalent of heat.
• If in a resistor, electrical energy is totally transformed into heat energy, the resistor is then called a passive resistor.
• Electrical power is to be transmitted from one place to another at high voltage and low current. This is the most important condition to minimize heat loss.
• If the potential difference at the two terminals of an electrical appliance is 1 volt and the current passing through it is 1 ampere, then the power of the appliance is l watt.
• Work done or electrical energy spent in 1 hour by an electrical appliance having power 1W is called 1Wh. Work done or electrical energy spent in lh by an electrical appliance having power lkW is called lkW-h or 1BOT unit.
• In writing 220V-100W on the body of an electric lamp or 220V-2000W on the body of an electric heater, both voltage rating and watt rating are preferred.
• If Q charge flows through a section under the potential difference V, then the amount of electrical work done,
• W = QV.
• The relation between joule (I), coulomb (C), and volt (V) is J = C x V [∵ W= QV]
• The heat evolved in a conductor of resistance R due to current I flowing through it for time t,
• \(H=\frac{I^2 R t}{J}\)
• where, J = mechanical equivalent of heat
• = 4.2 x 10⁷ erg/cal (In CGS)
• = 4.2 J/cal
• = 1 (In SI)
• i.e., \(H=0.24 I^2 R t=0.24 \frac{V^2 t}{R}\) [∵ V = IR]

• Current in a total circuit,
• I = \(\frac{E}{R+r}\) , where, E = emf of the cell, R = external resisR + r tance, r = internal resistance
• Power consumption in the circuit, \(P_0=I^2(R+r)=\frac{E^2}{R+r}\)
• Output power, \(P=I^2 R=\frac{E^2 R}{(R+r)^2}\)
• lW.h = 3600J, lkW.h = 3.6 x 106 J
• \(1 W \cdot h=\frac{W \times h}{1000}=\frac{V \times A \times h}{1000}\)
• For a parallel combination of a number of electrical appliances, equivalent power, P = P₁ + P₂ + P₃+….
• For a series combination of a number of electrical appliances, equivalent power,
• \(P^{\prime}=\frac{1}{\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots}\)
• If a heating coil of resistance R, consumes power P, when voltage V is applied to it, then by keeping V constant if it is cut in n equal parts then the resistance of each part will be R/n and the power consumed by each part P’ = nP.
• Power consumed by a n equal resistors is parallel is n₂ times that of power consumed in series if V remains the same i.e., P_p = n₂P_s.

Electric Energy And Power Very Short Questions and Answers

Question 1. A current I flows through a potential drop V across a conductor. What is the rate of heat production?
Answer: VI

Question 2. A small heating element connected to a 10 V dc supply draws a current of 5 A. Find the electric power supplied to the heater.
Answer: 50W

Question 3. A 220 V – 1000 W electric heater is connected in parallel with a 220 V – 60 W electric lamp, and the combination is fed by a 220 V main. Now, if the lamp is replaced by another 220 V – 100 W lamp, what will be the change in the rate of heat generation in the heater?
Answer: No change

Question 4. Two wires having resistances R and 2R are connected in series. If current is allowed to pass through the combination, what will be the ratio of power consumed in the two resistances?
Answer: 1: 2

Question 5. Two wires having resistances R and 2R are connected in parallel. If current is allowed to pass through the combination, what will be the ratio of power consumed in the two resistances?
Answer: 2: 1

Question 6. A 240 V- 1000 W lamp and a 240 V- 100 W lamp—which of these two has a thinner filament?
Answer: 240 V – 100 VV lamp

Question 7. Two resistances, each of magnitude 2Ω, are connected in series and a potential difference of 2V is applied at the two terminals of the combination. What is the power of the combination?
Answer: 1W

Question 8. Two resistances each of magnitude 2Ω are connected in parallel and a potential difference of 2V is applied at the two ends of the combination. What is the power of the combination?
Answer: 4W

Question 9. What is the resistance of an electric bulb marked 220 V-100 W in an incandescent state?
Answer: 484Ω

Question 10. An electric lamp is marked 240V -60 W. What is the resistance of the lamp in an incandescent state?
Answer: 960Ω

Question 11. A 220V -60 W electric lamp is connected to a 220V supply line. Determine the resistance of the lamp in an incandescent state.
Answer: 8.67Ω

Question 12. Which one of two electrical appliances, rated 100 W-200 V and 60 W-200 V, would have a higher resistance?
Answer: 60 W-220 V

Question 13. What is the largest voltage you can safely put across a 98Ω, 0.5 W resistor?
Answer: 7V

Electric Energy And Power Fill In The Blanks

1. A 2Ω resistance is connected to a source of constant emf. Another 2Ω resistance is connected in parallel to the previous one. The power consumed in the circuit becomes double

2. Two resistances are connected in series. If the current is made to pass through the combination, power consumed in the larger resistance will be higher

3. Two resistances are connected in parallel. If the current is made to pass through the combination power consumed in the larger resistance will be lower

4. A 220 V-100 W lamp and a 220V-60W lamp are connected in parallel. If the current is made to pass through the combination, the brightness of the first lamp will be higher

5. A 220 V-100 W lamp and a 220V-60W lamp are connected in series. If the current is made to pass through the combination, the brightness of the first lamp will be lower

Electric Energy And Power Assertion Reason Type

Direction: These questions have Statement 1 and Statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1
2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1
3. Statement 1 is true, Statement 2 is false
4. Statement 1 is false, Statement 2 is true

Question 1.

Statement 1: Electric current is distributed in different branches of a circuit in such a way, that the total heat evolved in the circuit is the lowest.

Statement 2: The transformation of electrical energy into heat energy in a circuit is less probable than its transformation into other forms of energy.

Answer: 3. Statement 1 is true, Statement 2 is false

Question 2.

Statement 1: An external circuit can draw a maximum power of 9 W from a source of emf 6 V and internal resistance 1 n.

Statement 2: The condition, for which an external circuit of resistance R draws the maximum power from a source of internal resistance r, is R = r.

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Question 3.

Statement 1: The power consumed would be 50 W by each of two 200V-100W lamps when their series combination is driven by a potential difference of 200V.

Statement 2: If P is the power consumed by a series combination of some electrical devices of power P₁, P₂, P₃,…., then, \(\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}+\cdots\)

Answer: 4. Statement 1 is false, Statement 2 is true

Question 4.

Statement 1: A fuse wire of diameter 0.5 mm can withstand a maximum current of I A. For a current of 8 A, a fuse wire made of the same alloy should have a diameter of 2 mm.

Statement 2: The radius r of a fuse wire and the maximum safe current I that may pass through it are related as \(I \propto r^{3 / 2}\)

Answer: 1. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Question 5.

Statement 1: The coil resistance of a 200 V-100 W electric fan is 20Ω. A power of 5 W is lost as heat when the fan rotates at its maximum speed.

Statement 2: If, for an electrical device the current is I and the terminal potential difference is V, the power consumed = VI

Answer: 2. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1

Electric Energy And Power Match The Columns

Question 1. Each of the two cells in the circuit has emf 5 V and internal resistance 2Ω. Match the following two columns for this circuit.

Answer: 1-B, 2-D, 3-A, 4-C

Question 2. The cell in the circuit has negligible internal resistance. The resistance values are in Ω. Match the following two columns for this circuit.

Answer: 1-C, 2-A, 3-D, 4-B

Question 3. The ratings of a few electrical elements are given in Column A, and Column B shows the values of maximum safe currents. Match the two columns.

Answer: 1-B, 2-A, 3-D, 4-C