NCERT Class 12 Physics — Current Electricity, Chapter 3 Exercises. All 9 questions solved.

The exercises at the end of Chapter 3 fall into four groups. Questions 3.1, 3.2 and 3.8 are about a cell with internal resistance — the most current it can give, the voltage at its terminals when it drives a current, and what changes when it is being charged. Questions 3.3 to 3.6 are about resistivity and how resistance rises with temperature. Question 3.7 is a network solved with Kirchhoff’s rules, and Question 3.9 is about the drift speed of electrons in a wire. The arithmetic runs on three results:

• A cell with internal resistance — a cell of emf $\varepsilon$ and internal resistance $r$ driving a current through an external resistance $R$ gives $I = \dfrac{\varepsilon}{R + r}$, and the voltage across its terminals is $V = \varepsilon – Ir$.
• Resistivity and temperature — $R = \dfrac{\rho l}{A}$, and over a moderate range $R_T = R_0\left[1 + \alpha(T – T_0)\right]$, where $R_0$ is the resistance at the reference temperature $T_0$ and $\alpha$ is the temperature coefficient.
• Current and drift speed — $I = neAv_d$, where $n$ is the number of free electrons per unit volume, $A$ the cross-section and $e = 1.6\times10^{-19}\ \text{C}$.

Key insight. Every question here is $V = IR$ applied to the right piece of the circuit. For a cell, that piece includes the cell’s own internal resistance: the terminal voltage is $\varepsilon – Ir$ when the cell drives the current (3.2) but $\varepsilon + Ir$ when an outside supply forces current backwards through it (3.8). For a heater, $V = IR$ at two moments gives two resistances, and the change between them gives the temperature (3.6). And Kirchhoff’s loop rule for a network is nothing more than $V = IR$ added up round a closed path (3.7).

Question 3.1

The storage battery of a car has an emf of $12\ \text{V}$. If the internal resistance of the battery is $0.4\ \Omega$, what is the maximum current that can be drawn from the battery?

Solution. The current is $I = \varepsilon/(R + r)$, which is largest when the external resistance $R$ is as small as it can be, that is, zero — the terminals joined by a thick wire. Then only the internal resistance limits the current:

$$I_{\max} = \frac{\varepsilon}{r} = \frac{12}{0.4} = 30\ \text{A}$$

Question 3.2

A battery of emf $10\ \text{V}$ and internal resistance $3\ \Omega$ is connected to a resistor. If the current in the circuit is $0.5\ \text{A}$, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Solution. The emf drives the current through the external resistor and the battery’s internal resistance together, so $I = \varepsilon/(R + r)$:

$$R + r = \frac{\varepsilon}{I} = \frac{10}{0.5} = 20\ \Omega \quad\Rightarrow\quad R = 20 – 3 = 17\ \Omega$$

Part of the emf is used up driving the current through the internal resistance, so the voltage available at the terminals is less than the emf:

$$V = \varepsilon – Ir = 10 – (0.5)(3) = 8.5\ \text{V}$$

As a check, this is also the voltage across the external resistor, $IR = (0.5)(17) = 8.5\ \text{V}$.

Question 3.3

At room temperature ($27.0\ ^\circ\text{C}$) the resistance of a heating element is $100\ \Omega$. What is the temperature of the element if the resistance is found to be $117\ \Omega$, given that the temperature coefficient of the material of the resistor is $1.70\times10^{-4}\ ^\circ\text{C}^{-1}$.

Solution. Take room temperature as the reference, so $R_0 = 100\ \Omega$ at $T_0 = 27.0\ ^\circ\text{C}$. The relation $R_T = R_0[1 + \alpha(T – T_0)]$ says the fractional rise in resistance is $\alpha$ times the rise in temperature:

$$T – T_0 = \frac{R_T – R_0}{R_0\,\alpha} = \frac{117 – 100}{100\times1.70\times10^{-4}} = \frac{17}{1.70\times10^{-2}} = 1000\ ^\circ\text{C}$$

That is the rise in temperature, so the element is at

$$T = 27 + 1000 = 1027\ ^\circ\text{C}$$

Question 3.4

A negligibly small current is passed through a wire of length $15\ \text{m}$ and uniform cross-section $6.0\times10^{-7}\ \text{m}^2$, and its resistance is measured to be $5.0\ \Omega$. What is the resistivity of the material at the temperature of the experiment?

Solution. The current is kept negligibly small so that it does not heat the wire; the resistance measured is then the resistance at the temperature of the surroundings. Resistivity is resistance scaled to a unit length and unit cross-section, so from $R = \rho l/A$:

$$\rho = \frac{RA}{l} = \frac{(5.0)(6.0\times10^{-7})}{15} = 2.0\times10^{-7}\ \Omega\ \text{m}$$

Question 3.5

A silver wire has a resistance of $2.1\ \Omega$ at $27.5\ ^\circ\text{C}$, and a resistance of $2.7\ \Omega$ at $100\ ^\circ\text{C}$. Determine the temperature coefficient of resistivity of silver.

Solution. Take the lower temperature as the reference, $R_0 = 2.1\ \Omega$ at $T_0 = 27.5\ ^\circ\text{C}$. The temperature coefficient is the fractional change in resistance per degree:

$$\alpha = \frac{R_T – R_0}{R_0\,(T – T_0)} = \frac{2.7 – 2.1}{2.1\times(100 – 27.5)} = \frac{0.6}{152.25} = 3.9\times10^{-3}\ ^\circ\text{C}^{-1}$$

The length and cross-section of the wire barely change with temperature, so the coefficient of resistance found this way is also the coefficient of resistivity.

Question 3.6

A heating element using nichrome connected to a $230\ \text{V}$ supply draws an initial current of $3.2\ \text{A}$ which settles after a few seconds to a steady value of $2.8\ \text{A}$. What is the steady temperature of the heating element if the room temperature is $27.0\ ^\circ\text{C}$? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70\times10^{-4}\ ^\circ\text{C}^{-1}$.

Solution. At the instant it is switched on the element has not yet warmed up, so the initial current gives its resistance at room temperature. The current then falls because the element heats up and its resistance rises; the steady current gives its resistance at the final temperature:

$$R_{27} = \frac{230}{3.2} = 71.9\ \Omega, \qquad R_T = \frac{230}{2.8} = 82.1\ \Omega$$

With room temperature as the reference, as in Question 3.3:

$$T – 27.0 = \frac{R_T – R_{27}}{R_{27}\,\alpha} = \frac{82.1 – 71.9}{71.9\times1.70\times10^{-4}} = 840\ ^\circ\text{C}$$

so $T = 27 + 840 = 867\ ^\circ\text{C}$.

Question 3.7

Determine the current in each branch of the network shown in Fig. 3.20:

Solution. The book’s figure leaves the lower junction unlabelled; the answer key calls it $D$, and so does this solution. The network is a bridge, but not a balanced one — $AB : BC = 10 : 5$ while $AD : DC = 5 : 10$ — so current does flow in $BD$ and it has to be found with Kirchhoff’s rules. The junction rule (charge is conserved at a junction) lets three unknown currents describe all six branches. Call the current in $AB$ $I_1$, in $AD$ $I_2$, and in $BD$ (from $B$ to $D$, the direction of the arrow in the figure) $I_3$. Then

$$I_{BC} = I_1 – I_3,\qquad I_{DC} = I_2 + I_3,\qquad I = I_1 + I_2 \ \text{(through the cell)}$$

The loop rule (the potential drops round any closed loop add up to the emfs in it) gives three equations.

Loop ABDA, which contains no cell:

$$10I_1 + 5I_3 – 5I_2 = 0 \quad\Rightarrow\quad I_2 = 2I_1 + I_3$$

Loop BCDB, which contains no cell:

$$5(I_1 – I_3) – 10(I_2 + I_3) – 5I_3 = 0 \quad\Rightarrow\quad I_1 – 2I_2 – 4I_3 = 0$$

Loop ABC and back through the cell and the $10\ \Omega$ resistor to A. The cell’s long (positive) plate faces the $10\ \Omega$ resistor, so it drives current round this loop in the direction taken:

$$10I_1 + 5(I_1 – I_3) + 10(I_1 + I_2) = 10 \quad\Rightarrow\quad 25I_1 + 10I_2 – 5I_3 = 10$$

Substituting $I_2 = 2I_1 + I_3$ into the second equation gives $-3I_1 – 6I_3 = 0$, so $I_3 = -\tfrac12 I_1$ and then $I_2 = \tfrac32 I_1$. Putting both into the third:

$$25I_1 + 15I_1 + 2.5I_1 = 10 \quad\Rightarrow\quad I_1 = \frac{10}{42.5} = \frac{4}{17}\ \text{A}$$

so $I_2 = \dfrac{6}{17}\ \text{A}$, $I_3 = -\dfrac{2}{17}\ \text{A}$, $I_{BC} = I_1 – I_3 = \dfrac{6}{17}\ \text{A}$, $I_{DC} = I_2 + I_3 = \dfrac{4}{17}\ \text{A}$ and $I = \dfrac{10}{17}\ \text{A}$.

The negative sign on $I_3$ is not an error: it says the current in $BD$ flows the other way from the arrow in the figure, from $D$ up to $B$. It makes sense, because $D$ is at a higher potential than $B$ — the drop from $A$ to $D$ is $5\times\tfrac{6}{17} = \tfrac{30}{17}\ \text{V}$, less than the drop from $A$ to $B$, $10\times\tfrac{4}{17} = \tfrac{40}{17}\ \text{V}$.

As a check, the junction rule holds at $B$ ($\tfrac{4}{17} + \tfrac{2}{17}$ in, $\tfrac{6}{17}$ out) and at $D$ ($\tfrac{6}{17}$ in, $\tfrac{2}{17} + \tfrac{4}{17}$ out), and the bridge as a whole behaves as a $7\ \Omega$ resistor: $A$ is $\tfrac{70}{17}\ \text{V}$ above $C$ for a current of $\tfrac{10}{17}\ \text{A}$. With the $10\ \Omega$ in series the circuit totals $17\ \Omega$, and $10\ \text{V}/17\ \Omega = \tfrac{10}{17}\ \text{A}$.

Question 3.8

A storage battery of emf $8.0\ \text{V}$ and internal resistance $0.5\ \Omega$ is being charged by a $120\ \text{V}$ dc supply using a series resistor of $15.5\ \Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Solution. In the charging circuit the $120\ \text{V}$ supply and the $8.0\ \text{V}$ battery are connected positive to positive, so their emfs oppose each other and the net emf driving the current is $120 – 8 = 112\ \text{V}$. The whole loop resistance is the series resistor plus the battery’s internal resistance:

$$I = \frac{120 – 8.0}{15.5 + 0.5} = \frac{112}{16} = 7.0\ \text{A}$$

The current is forced into the battery’s positive terminal, against its emf. So across the battery’s terminals there is its emf plus the drop across its internal resistance, not minus:

$$V = \varepsilon + Ir = 8.0 + (7.0)(0.5) = 11.5\ \text{V}$$

The same value comes from the supply’s side: $120 – (7.0)(15.5) = 120 – 108.5 = 11.5\ \text{V}$.

Without the series resistor the only resistance in the loop would be the battery’s $0.5\ \Omega$, and the current would be $112/0.5 = 224\ \text{A}$ — enough to overheat and damage the battery and the wiring. The series resistor limits the charging current to a safe value.

Question 3.9

The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5\times10^{28}\ \text{m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0\ \text{m}$ long to its other end? The area of cross-section of the wire is $2.0\times10^{-6}\ \text{m}^2$ and it is carrying a current of $3.0\ \text{A}$.

Solution. The current is the charge carried past a cross-section each second by the drifting electrons, $I = neAv_d$, so the drift speed is

$$v_d = \frac{I}{neA} = \frac{3.0}{(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})} = 1.1\times10^{-4}\ \text{m s}^{-1}$$

At that speed an electron takes

$$t = \frac{l}{v_d} = \frac{3.0}{1.1\times10^{-4}} = 2.7\times10^4\ \text{s}$$

which is about $7.5$ hours to cover $3\ \text{m}$. A lamp still lights at once when switched on because the electric field is set up along the whole wire almost instantly, and electrons everywhere in the wire start drifting together; no single electron has to travel from the switch to the lamp.

Note on the book’s answer. The key prints $2.7\times10^4\ \text{s}$ ($7.5\ \text{h}$). Unrounded, the time is $27{,}200\ \text{s}$, or $7.6\ \text{h}$; the key’s hours are the rounded seconds divided by $3600$. This is a rounding difference only.

Common mistakes

• Question 3.1: using some external resistance. The maximum current is drawn when the external resistance is zero, so only the internal resistance limits it. Students who look for an $R$ in the question, or think the current can be unlimited, have forgotten that the cell resists its own current.
• Question 3.2: forgetting the internal resistance. Dividing $10\ \text{V}$ by $0.5\ \text{A}$ gives $20\ \Omega$, which is the resistance of the whole circuit, battery included. The resistor is $17\ \Omega$, and for the same reason the terminal voltage is $8.5\ \text{V}$, not $10\ \text{V}$.
• Question 3.3: giving $1000\ ^\circ\text{C}$. The formula gives the rise in temperature above the reference temperature, $27\ ^\circ\text{C}$, so the room temperature must be added back to get $1027\ ^\circ\text{C}$.
• Question 3.6: dividing by the hot resistance. $\alpha$ is defined relative to the resistance at the reference temperature, so the change $R_T – R_{27}$ is divided by the room-temperature value $71.9\ \Omega$, which comes from the initial current, not by $82.1\ \Omega$.
• Question 3.7: treating the bridge as balanced. It looks like a Wheatstone bridge, so students put zero current in $BD$. The balance condition fails here ($10/5 \neq 5/10$), and $\tfrac{2}{17}\ \text{A}$ flows in $BD$.
• Question 3.7: reading a negative current as a mistake. The direction of each unknown current is a guess made before solving. A negative answer, as for $BD$, only means the current flows the other way; the size is still right.
• Question 3.8: writing $\varepsilon – Ir = 4.5\ \text{V}$. That is the terminal voltage of a battery delivering current. Here current is driven into the positive terminal to charge it, so the internal drop adds to the emf, giving $11.5\ \text{V}$.
• Question 3.9: expecting the electrons to be fast. A drift speed of about $0.1\ \text{mm}$ per second looks wrong, so students hunt for a missing power of ten. It is right: current is large because $n$ is enormous, not because the electrons move quickly.

Practise next

• Chapter 4, Moving Charges and Magnetism — the currents of this chapter produce magnetic fields and feel forces in them.
• Chapter 2, Electrostatic Potential and Capacitance — the potential difference that drives every current here, and the capacitors that store charge instead of passing it.