## Current Electricity

## Electric Current and Ohm’s Law Short Question And Answers

**Question 1. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27 °C? The temperature coefficient of resistance of nichrome averages over the temperature range involved is 1.70 x 10 ^{-4}ºCT_{1}.**

**Answer:**

The temperature coefficient of resistance is given by

⇒ \(\alpha=\frac{R_2-R_1}{R_1\left(t_2-t_1\right)}\)

∴ \(t_2-t_1=\frac{R_2-R_1}{R_1 \alpha}=\frac{\frac{230}{2.8}-\frac{230}{\angle 3.2}}{\frac{230}{3.2} \times 1.8^{\top} \times 10^{-4}}\)

= 840

∴ t_{2} = (840 + 27) °C

= 867 ºC

**Question 2. A storage battery of emf 8.0 V and internal resistance 0.5 XI is being charged by a 120 V dc supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging? What is the purpose of using the series resistor in the charging circuit?**

**Answer:**

During charging,

V =E-I(R+ r)

⇒ \(I=\frac{E-V}{R+r}\)

= \(\frac{120-8}{15.5+0.5}\)

= 7A

∴ Terminal voltage of the battery

= V + Ir

= 8 + 7 x 0.5

= 11.5 V

The series resistor prevents the charging current from attaining a very value

**Question 3. The earth’s surface has a negative surface.qhafge density of 10 ^{-9} C.m^{-2}. The potential difference of 400kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in only 1800 A over the entire globe. If ‘there were no mechanism for sustaining an atmospheric electric field, how much time (roughly) would be required to neutralize the earth’s surface? [Radius of earth = 6.37 x 10^{6} m]**

**Answer:**

Surface area of earth = 47T x (6.37 x 10^{6})^{2} m^{2}

The total charge on the surface of the earth

= area x surface density of charge

= 4π x (6.37 x 10^{-6})^{2} x 10^{-9} C

∴ Time taken for discharge, t = \(\frac{Q}{I}\)

∴ \(t=\frac{4 \pi \times 6.37 \times 6.37 \times 10^3}{1800}\)

= 283 s

**Question 4. Two wires of equal length, one of aluminum and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. [\(\rho_{\mathrm{Al}}=2.63 \times 1 \mathrm{Q}^{-8} \Omega \cdot \mathrm{m}, \quad \rho_{\mathrm{Cu}}=1.72 \times 10^{-8} \Omega \cdot \mathrm{m}\) relative density of A _{1} = 2.7, of Cu = 8.9 ]**

**Answer:**

Mass of the wire, M = A.l.d

[where A = area of cross-section of the wire, l = length of the

wire, d = density of the material of the wire]

∴ A = \(\frac{M}{ld}\)

Resistance of the wire, \(R=\rho \cdot \frac{l}{A}=\rho \cdot \frac{l^2}{M} d\)

∵ \(R_{\mathrm{Al}}=R_{\mathrm{Cu}}\) and l is same for both the wires,

⇒ \(\frac{\rho_{\mathrm{Al}} \cdot d_{\mathrm{Al}}}{M_{\mathrm{Al}}}=\frac{\rho_{\mathrm{Cu}} \cdot d_{\mathrm{Cu}}}{M_{\mathrm{Al}}}\)

∴ \(\frac{M_{\mathrm{Cu}}}{M_{\mathrm{Al}}}=\frac{\rho_{\mathrm{Cu}}}{\rho_{\mathrm{Al}}} \times \frac{d_{\mathrm{Cu}}}{d_{\mathrm{Al}}}=\frac{1.72 \times 10^{-8}}{2.63 \times 10^{-8}} \times \frac{8.9}{2.7}\)

∴ The copper wire is 2.16 times heavier than the aluminum wire i.e., aluminium wire is lighter. This is the reason for using aluminum wire in overhead power cables.

**Question 5. The length, diameter, and specific resistance of two wires of different materials are each in the ratio 2:1. One of the wires has a resistance of 10 ohms. Find the resistance of the other wire**

**Answer:**

⇒ \(R=\rho \frac{l}{A}=\rho \frac{l}{\pi d^2 / 4}\)

Now, for the two wires

⇒ \(\frac{R_1}{R_2}=\frac{\rho_1}{\rho_2} \cdot \frac{l_1}{l_2}\left(\frac{d_2}{d_1}\right)^2=\frac{2}{1} \times \frac{2}{1} \times\left(\frac{1}{2}\right)^2\)

= 1

Hence, if one of the wires has a resistance of 10Ω, the resistance of the other wire must be 10Ω.

**Question 6. Draw a graph representing the change in specific resistance with temperature.**

**Answer:**

Let a’ be the coefficient of linear expansion and a be the temperature coefficient of resistance of the material of the conductor.

If the resistivity of the material of the conductor at 0°C and t°C are ρ_{0} and p respectively then,

ρ = ρ_{0}[1 + (α + α’)f]

or, ρ = ρ_{0}(α + α’)t + ρ_{0}

It is similar to the general equation of a straight line i.e.,

y = mx + c

**Question 7. Find the equivalent resistance between the two ends A and B of the following circuit**

The equivalent circuit of the given circuit.

The resistance of each resistor = R

Equivalent resistance between A and C = \(\frac{R}{3}\)

Equivalent resistance between C and B = \(\frac{R}{3}\)

∴ The equivalent resistance between A and B \(=\frac{R}{3}+\frac{R}{3}=\frac{2 R}{3}\)

**Question 8. Define lost volt. State the factors on which the internal resistance of a cell depends**

**Answer:**

If E is the emf of a given cell and V is the potential difference across a given circuit then, V = E-Ir, i.e., the whole emf of the cell is not obtained as a potential difference in the external circuit.

A potential of magnitude IT is lost inside the cell for an internal resistance of r. Thisis called thelostvoltofthe cell.

The internal resistance of a cell depends on the size of two electrodes, the distance between two electrodes, and the nature of the electrolyte.

**Question 9. Of ammeter and voltmeter whose resistance is greater and why?**

**Answer:**

The resistance of the voltmeter is greater than that of the ammeter. A voltmeter is connected in parallel with the main circuit and has higher resistance so that it does not change the current in the main circuit.

**Question 10. **

**1. What will be the charge on the capacitor in the circuit given below?**

**2. Find the energy storedin the capacitor**

**Answer:**

1. In a dc circuit, the current through a capacitor is zero. The voltage across the capacitor= voltage across the 5Ω

resistor = \(3 \times \frac{5}{5+10}=1 \mathrm{~V}\)

∴ Charge of the capacitor,

Q = CV

= 10μF X 1V

= 10μC

= 1 X 10^{-5}C

2. Energy storedin the capacitor

⇒ \(\frac{1}{2} C V^2=\frac{1}{2} \times 10 \mu \mathrm{F} \times(1 \mathrm{~V})^2\)

= 5μJ

= 5 X 10^{-6} J

**Question 11. Two cells of emf E _{1}, E_{2}, and internal resistances r_{1}, r_{2} respectively are connected in parallel combination. Determine the equivalent of the combination.**

**Answer:**

If the equivalent emf of the combination is E_{0} and its equivalent internal resistance is r_{0}, then

⇒ \(r_0=\frac{r_1 r_2}{r_1+r_2}, i_1=\frac{E_1}{r_1}, i_2=\frac{E_2}{r_2} \text { and } i=\frac{E_0}{r_0}\)

∵ \(i=i_1+i_2, \frac{E_0}{r_0}=\frac{E_1}{r_1}+\frac{E_2}{r_2}\)

or, \(E_0=E_1 \frac{r_0}{r_1}+E_2 \frac{r_0}{r_2}=E_1 \frac{r_2}{r_1+r_2}+E_2 \frac{r_1}{r_1+r_2}\)

⇒ \(\frac{1}{r_1+r_2}\left(E_1 r_2+E_2 r_1\right)\)

**Question 12. Estimate the average drift velocity of conduction electrons in a copper wire of cross section 2.0 x 10 ^{-3}cm^{2} carrying a current of 2.0A. Assume the density of conduction electrons to be 9 x 10^{28}m^{-3}**

**Answer:**

⇒ \(v_d=\frac{I}{n e A}\)

= \(\frac{2.0}{\left(9 \times 10^{28}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(2.0 \times 10^{-7}\right)}\)

= 6.94 X 10^{-4}m.s^{-1} (approx.)

**Question 13. Under what condition will the terminal potential difference be more than the emf of a cell?**

**Answer:**

V = E-Ir, if I is negative, then V > E, i.e., the terminal potential difference of the cell is more than its emf. This occurs if other cells in the circuit cause current to flow through the given cell from its positive electrode to the negative electrode, which is opposite to the direction in which current flows in a cell.

**Question 14. Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?**

**Answer:**

⇒ \(R=\rho \frac{l}{A}\)

Here, R and l are both equal for the copper and manganin wires.

∴ \(R=\rho_1 \frac{l}{A_1}\)

= \(\rho_2 \frac{l}{A_2} \quad\)

or, \(\frac{A_1}{A_2}=\frac{\rho_1}{\rho_2}\)

The resistivity of copper (Pj) is less than that (p2) of manganin.

∴ A_{1} < A_{2} i.e., the area of cross section of the manganin wire (A_{2}) is greater. So the manganin wire is thicker.

**Question 15. Define the relaxation time of the free electrons drifting in the conductor. How is it related to the drift velocity of free electrons? Use tills relation to deduce the expression for the electrical resistivity of the material.**

**Answer:**

The relaxation time or mean free time (t0) of a free electron Inside a conductor is defined as the average time spent by the electron between two successive collisions.

It is assumed that just after a collision, the electron velocity is reduced to zero. Then the force on it due to the applied electric field E is eE.

So the acceleration of the electron is \(\frac{eE}{m}\) and the velocity attained by it just before the next collision = \(\frac{eE}{m}\)t_{0}. Therefore, the average or drift velocity of free electrons inside a conductor is

⇒ \(v=\frac{0+\frac{e E}{m} t_0}{2} \quad \text { or, } v=\frac{e E}{2 m} t_0\)

or, \(t_0=\frac{2 m}{e E} v\)…..(1)

We also know, \(v=\frac{I}{n e A}\)

So, \(\frac{I}{n e A}=\frac{e E}{2 m} t_0 \quad \text { or, } E=\frac{2 m}{n e^2 A t_0} I\)

Now, if the potential difference is applied between the ends of a conductor of length 1, then E = \(\frac{V}{l}\)

Then, its resistance, \(\mathrm{R}=\frac{V}{I}=\frac{E l}{I}=\frac{2 m l}{n e^2 A t_0}=\rho \frac{l}{A}\)

So, the electrical resistivity of the material is

⇒ \(\rho=\frac{2 m}{n e^2 t_0}\)…..(2)

Also, by substituting t_{0} from (1), we have

⇒ \(\rho=\frac{e E}{n e^2 v}=\frac{E}{n e v}\)….(3)

**Question 16. A cell of emf E and internal resistance r is connected across a variable resistor R. Plot a graph showing variation terminal voltage V of the of the cell versus the current. Using the plot, show how the emf of the cell and its internal resistance can be determined.**

**Answer:**

Here,emf= terminal voltage + internal potential drop

So, E = V + Ir,

or, V = -Ir + E

This equation is of the form y = mx + c.

So the I-V graph is a straight line of slope -r and of intercept E on the V-axis

The emf E is known from this intercept and the internal resistance r from the slope i.e., r = – \(\frac{AB}{BC}\).

**Question 17. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10 ^{-7} m^{2} carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 10^{28} m^{-3}.**

**Answer:**

Average drift speed,

⇒ \(\nu=\frac{I}{n e A}=\frac{1.5}{\left(9 \times 10^{28}\right) \times\left(1.6 \times 10^{-19}\right) \times\left(1.0 \times 10^{-7}\right)}\)

= 1.04 X 10^{3} m.s^{-1}

**Question 18. Two metallic resistors are connected first in series and then in parallel across a dc supply. The plot of I – V graph is shown for the two cases. Which one represents a parallel combination of the resistors and why?**

Graph A represents a parallel combination of resistors since

⇒ \(\frac{d V}{d I}=R \quad \text { or, } \frac{d I}{d V}=\frac{1}{R}\)

A parallel combination has low resistance, and hence the

graph will have a greater slope.

**Question 19. Two identical cells of emf 1.5 Veach joined in parallel supply energy to an external circuit consisting of two resistances of 7Ω each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4 V. Calculate the internal resistance of each cell.**

**Answer:**

The two cells are connected in parallel. So, the equivalent emf is 1.5 V.

Now, the two resistors are connected in parallel. So, the equivalent resistance is

⇒ \(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}\)

∴ \(R_{\mathrm{eq}}=\frac{R}{2}=\frac{7}{2}=3.5 \Omega\)

The terminal voltage of the cells measured by the voltmeter is 1.4 V.

The equivalent internal resistance of the combination of cells is,

⇒ \(r_{\mathrm{eq}}=\left(\frac{E-V}{V}\right) R\)

∴ \(r_{\text {eq }}=\frac{1.5-1.4}{1.4} \times 3.5=\frac{0.1}{1.4} \times 3.5=0.25 \Omega\)

As the cells are connected in parallel,

So, \(r_{\text {eq }}=\frac{r^{\prime}}{2} \quad\left[r^{\prime}=\text { internal resistance of each cell }\right]\)

∴ r’ = 2r_{eq}

= 2 x 0.25

= 0.50

**Question 20. A wire whose cross-sectional area is increasing linearly from one end to the other, is connected across a battery of V volts. Which of the following quantities remain constant in the wire?**

**Drift speed****Current density****Electric current****Electric field**

**Answer:**

The electric current remains constant in the wire so that no charge accumulates in the wire. The drift speed and current density depend on the area of cross, section and hence they do not remain constant. From the relation, j = σE, it can be deduced that the electric field does not remain constant.

**Question 21. An ammeter A and a resistor of 4Ω are connected to the terminals of the source. The emf of the source is 12V having an internal resistance of 2Ω. Calculate the voltmeter and ammeter readings.
Answer:
**

The ammeter reading is, \(I=\frac{E}{R+r}\)

= \(\frac{12}{4+2}\)

= \(2 \mathrm{~A}\)

= 2A

The voltmeter reading is V = E-Ir

= 12 – 2 x 2

= 8V

**Question 22. **

**1. Define the term ‘conductivity’ of an electric wire. Write its SI unit.**

**2. Using the concept of free electrons In A conductor, derive thcrmAnpislon for the conductivity of a wire in terms of number density timer. Ifence obtains the ratio between current density and the applied electric field E.**

**Answer:**

1. The conductivity of a mental wins lu the ratio of the electrical current density to the applied electric field,

Electrical conductivity, \(\sigma=\frac{l}{D}\)

The unit of conductivity is Ω^{-1}.m^{-1}

2. The electric field E exerts an electric force on an electron, \(\vec{F}\) = – \(\vec{E}\)e

Acceleration of each electron, \(\vec{a}=\frac{-e \vec{E}}{m}\)…(1)

where m = mass of an electron and e = charge on an electron.

Drift velocity,

⇒ \(\vec{v}_d=\frac{\vec{v}_1+\vec{v}_2+\cdots+\vec{v}_n^{\prime}}{n}\)

⇒ \(\frac{\left(\vec{u}_1+\vec{a} \tau_1\right)+\left(\vec{u}_2^{\prime}+\vec{a} \tau_2\right)+\cdots \cdot+\left(\vec{u}_n+\vec{a} \tau_n\right)}{n}\)

Now, \(\vec{u}_1, \vec{u}_2, \cdots, \vec{u}_n\) are the thermal Velocities of the

electrons,

⇒ \(\vec{a} \tau_1, \vec{a} \tau_2, \cdots, \vec{a} \tau_n\) are the velocities acquired by electrons,

⇒ \(\tau_1, \tau_2, \cdots, \tau_n\) are the time elapsed after the collision

∴ \(v_d=\frac{\left(\vec{u}_1+\vec{u}_2+\cdots+\vec{u}_n\right)}{n}+\frac{\vec{a}\left(\tau_1+\tau_2+\cdots+\tau_n\right)}{n}\)

Since \(\frac{\vec{u}_1+\vec{u}_2+\cdots+\vec{u}_n}{n}\) average thermal velocity = 0, we get

⇒ \(\vec{v}_d=\vec{a} \tau\)….(2)

where \(\tau=\frac{\tau_1+\tau_2+\tau_3+\cdots+\tau_n}{n}\) is the average time elapsed between two successive collisions of electrons which is known as relaxation time of electron

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

⇒ \(\vec{v}_a=\frac{-e \vec{E}}{m} \tau\)….(3)

Let N number of free electrons pass through the cross-section (A) of the wire In time t, Is confined within a cylinder of length l. Then electric current flows through the conductor,

⇒ \(I=\frac{-N e}{t}=\frac{-n A l e}{\frac{l}{v_d}}\) [n = number density of free electrons)

or, \(I=-n e A v_d=\frac{n e^2 A \tau}{m} E\)

or, \(J=\frac{I}{A}=\left(\frac{n e^2 \tau}{m}\right) E=\sigma R\)

**Question 23. A 10 V cell of negligible internal resistance Is connected In parallel across a battery of emf 200 V and internal resistance 38Ω and the value of current In the circuit.**

**Answer:**

The current in the circuit,

⇒ \(I=\frac{\text { effective emf }}{\text { resistance }}=\frac{200-10}{38}=5 \mathrm{~A}\)