WBCHSE Class 12 Physics Electric Energy And Power Questions And Answers

WBCHSE Class 12 Physics Electric Energy And Power Questions And Answers

Question 1. Among emf, energy, power, and charge, which one has the unit As?
Answer:

A.s = unit of current x unit of time

By definition, the charge (Q) flowing through a conductor in unit time, is called current (I).

∵ I = \(\frac{Q}{t}\)

or, Q = It

So, charge = current x time

∴ A.s is the unit of charge.

Question 2. How are lamps, fans, etc. connected in domestic electrical connection in series combination or parallel combination?
Answer:

The voltage rating of the lamp, fan, etc. is generally fixed at a definite value (for example, 220 V ), and at this voltage, they work at the maximum rate. Again, we know that if a few conductors are connected in parallel, the potential difference across each conductor is equal. So, in domestic electric wiring lamps, fans, etc. are connected in parallel combination.

Question 3. The speed of an electric fan is reduced with the help of a regulator. What will happen in the energy consumption?
Answer:

If an electric fan of resistance R runs under potential difference V for time t, then work done,

⇒ \(W=I^2 R t=\frac{I^2 R^2 t}{R}=\frac{V^2 t}{R}\)

To reduce the speed, some resistance, say R1 is added in series through a regulator. So the potential difference at the two ends of the combination of R and R1 is V.

So, work done in time t, \(W_1=\frac{V^2 t}{R+R_1}\)

Evidently, Wx is less than W. So, energy consumption decreases.

Question 4. Is the filament of the lamp marked ‘240 V-1000 W’ thin or thick in comparison to the filament of the lamp marked ‘240 V-100W’?
Answer:

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

So, if the same potential difference is applied to both lamps, the resistance of the lamp having larger power will be less than that of the lamp having less power.

Again, the larger the cross-sectional area (A) of a conducting wire (filament), the smaller will be its resistance (since \(R \propto \frac{1}{A}\) ). So, the filament of a 1000 W lamp will be thicker than that of a 100 W lamp.

Question 5. A series combination of a 60 W and a 100 W lamp is connected to the mains. Which lamp will glow brighter and why?
Answer:

Resistance of 60 W lamp, \(R_1=\frac{V^2}{60}\);

V is the mains voltage.

Resistance of 100 W lamp, \(R_2=\frac{V^2}{100}\)

Obviously, R1 > R2.

As the two lamps connected in series, are connected to the mains, the same current will flow through them. So from Joule’s law [latex]H \propto I^2 R t[/latex], it can be said that more heat will be produced in the lamp having higher resistance i.e., 60 W lamp. Hence, the 60 W lamp will glow brighter than the 100 W lamp.

WBCHSE Class 12 Physics Electric Energy And Power Question And Answers

Question 6. A heater coil is cut into two equal parts and only one part is now used in the heater. What is the percentage of increase or decrease in the rate of production of heat?
Answer:

In both cases, the potential differences between the two ends of the coil (V) are the same. So, rate of production of heat \(\propto \frac{V^2}{R}\). When the coil is cut into two equal parts, the value of R becomes half. So, the rate of production of heat becomes double, i.e., this rate increases by 100%

Question 7. A few wires of the same dimension but of different specific resistances are connected in parallel and then this parallel combination is connected to a battery. In which wire will the rate of production of heat due to the Joule effect be maximum?
Answer:

As the wires are connected in parallel, so potential difference across each wire will be the same.

Heat produced, \(H=\frac{V^2 t}{R J}=\frac{V^2 t}{J \cdot \rho \frac{l}{A}}\)

As each wire is of the same dimension, the rate of production of heat will be maximum in the wire, whose specific resistance is minimum.

Question 8. The specific resistance of the material of a conducting wire is p and the current through unit cross-sectional area of the wire (i.e., current density) is J. What is the power consumed per unit volume of the wire?
Answer:

Let the length of the wire = l, the cross-sectional area of the wire = A, and current through the wire = I.

So, resistance, \(R=\rho \cdot \frac{l}{A}\) ;

current density, \(J=\frac{I}{A}\) ; volume of the

wire = lA

Power consumed, P = I²R

So, power consumed per unit volume,

⇒ \(\frac{P}{l A}=\frac{I^2 R}{l A}=\frac{I^2}{l A} \cdot \rho \frac{l}{A}=\left(\frac{I}{A}\right)^2 \cdot \rho=J^2 \rho\)

Question 9. A few electric bulbs are connected in series to the 220 V mains. One bulb is fused, and the remaining bulbs are again put in series and connected to the same supply of 220 V. In which case will the bulbs glow brighter and why?
Answer:

In the second case, the bulbs will glow brighter. Let n be the number of bulbs and the bulbs be connected to 220 V mains.

So, potential difference across each bulb, V1 = \(\frac{220}{n}\)V.

One bulb out of them is fused. So, now the potential difference across each bulb,

⇒ \(V_2=\frac{220}{n-1} \mathrm{~V} \quad ∴ V_2>V_1\)

So, in the second case, the bulbs will glow brighter.

Question 10. Three resistances of equal value are connected in four different combinations. Arrange them in increasing order of power dissipation.

Electric Energy And Power Three resistances of equal value

Answer:

Let R be the value of each resistance.

The equivalent resistance of the circuit (a), R1 = 3R

Equivalent resistance of circuit (b), R2 = \(\frac{R}{3}\)

The equivalent resistance of the circuit (c), \(R_3=\frac{2 R \cdot R}{2 R+R}=\frac{2}{3} R\)

The equivalent resistance of the circuit (d), \(R_4=\frac{R}{2}+R=\frac{3}{2} R\)

∴ R2 < R3 < R4 < R1

Since, power, P = I²R

So, P2 < P3 < P4 < P1.

Question 11. A heater coil has been cut into two equal parts and one coil is used as a heater. What is the percentage change in heat generation?
Answer:

The potential difference across the coU in both cases is V.

Therefore rate of heat generation = \(\frac{V^2}{R}\)

If the coil is cut into half then the value of R becomes half. That’s why, the rate of heat generation has doubled.

That is, this rate increases by 100%.

Question 12. Two electric bulbs of 50 W and 100 W are connected in mains once in

  1. Series and next in
  2. Parallel combination. Which bulb will glow brighter in each case?

Answer:

1. In a series combination, a 50 W bulb will glow brighter than a 100 W bulb. We know, the resistance of the bulb,

⇒ \(R=\frac{V^2}{P}\) [V = potential difference, P = power]

So the resistance of the 50W bulb is greater than that of the 100W bulb.

In this combination, the current in both bulbs is are same.

As heat generation is proportional to I²R, that’s why a 50W bulb glows brighter than a 100W bulb.

2. In parallel combination, a 100W bulb will glow brighter than a 50W bulb

When the bulbs are connected in parallel combination potential difference across each one is the same.

As, heat generation is proportional to \(\frac{V^2}{R}\), that’s why a 100 W bulb glows brighter than a 50 W bulb in parallel combination.

Question 13. A wire, when connected to a 220 V mains supply, has power dissipation. Now the wire is cut into two equal pieces, which are connected in parallel to the same supply. Power dissipation in this case is P2. What is the ratio of P2 and P1?
Answer:

If R is the resistance of the wire, then the resistance of each of the two pieces is

⇒ \(\frac{R}{2}\).

Equivalent resistance of the parallel combination,

⇒ \(r=\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}=\frac{R}{4} \quad \text { or, } \frac{R}{r}=4\)

So, \(\frac{P_2}{P_1}=\frac{\frac{V^2}{r}}{\frac{V^2}{R}}=\frac{R}{r}=4\)

Question 14. Power consumed in resistance R3 is P3. Determine the power consumed in resistances R1 and R2.

Electric Energy And Power Power consumed in resistance

Answer:

Let I1 and I2 be the currents flowing in the upward branch and downward branch respectively.

So, \(I_1\left(R_1+R_2\right)=I_2 R_3 \quad \text { or, } \frac{I_1}{I_2}=\frac{R_3}{R_1+R_2}\)

Now, \(\frac{P_1}{P_3}=\frac{I_1^2 R_1}{I_2^2 R_3}=\left(\frac{R_3}{R_1+R_2}\right)^2 \cdot \frac{R_1}{R_3}=\frac{R_1 R_3}{\left(R_1+R_2\right)^2}\)

or, \(P_1=P_3 \cdot \frac{R_1 R_3}{\left(R_1+R_2\right)^2}\)

Similarly \(P_2=P_3 \cdot \frac{R_2 R_3}{\left(R_1+R_2\right)^2}\)

Question 15. A generating station supplies electric power P at voltage V to a factory through a cable of resistance R. Show that the loss of power in the connecting cable is inversely proportional to V2.
Answer:

Supplied power, P = VI

∴ I = \(\frac{P}{V}\)

Loss in power, \(\Delta P=I^2 R=\frac{P^2 R}{V^2}\)

∴ \(\Delta P \propto \frac{1}{V^2}\)

Question 16. Prove that total heat produced in different resistors of the circuit is minimal when the current is divided into a number of branches.
Answer:

When the current is divided into a number of parallel branches, then I = I1 + I2

Heat produced in the circuit in time t,

⇒ \(H=\frac{I_1^2 r_1 t}{J}+\frac{\left(I-I_1\right)^2 r_2 t}{J}\)

Electric Energy And Power total heat produced

Differentiating both sides with respect to I1,

⇒ \(\frac{d H}{d I_1}=\frac{2 I_1 r_1 t}{J}-\frac{2\left(I-I_1\right) r_2 t}{J}\)

When the heat generated in the circuit is minimal,

⇒ \(\frac{d H}{d I_1}=0 \text { or, } \frac{2 I_1 r_1 t}{J}-\frac{2\left(I-I_1\right) r_2 t}{J}=0\)

or, \(\frac{2 I_1 r_1 t}{J}=\frac{2\left(I-I_1\right) r_2 t}{J} \text { or, } I_1 r_1=\left(I-I_1\right) r_2=I_2 r_2\)

∴ \(\frac{I_1}{I_2}=\frac{r_2}{r_1}\)

i.e., the heat produced in the circuit will be minimal if the current is divided into the branches such that the current in each branch is inversely proportional to the resistance.

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