NCERT Exemplar Solutions Class 12 Physics Current Electricity — Very Short, Short and Long Answer Questions Q12–31

NCERT Exemplar Problems, Class 12 Physics — Current Electricity, Very Short Answer Questions 12 to 21, Short Answer Questions 22 to 27 and Long Answer Questions 28 to 31. All 20 questions solved. The book numbers them 3.12 to 3.31.

Three ideas carry this page. Ohm’s law and resistivity come from collisions: the relaxation time $\tau$ fixes drift velocity, conductivity and the way resistance changes with temperature. A real cell is an emf in series with an internal resistance, so its terminal voltage is $E – Ir$ and everything about batteries in series, in parallel or in opposition follows from that. A null method needs no current from the source being measured, which is why the Wheatstone bridge and the potentiometer are accurate.

Key insight. Draw the circuit as emfs and resistances, write one loop equation, and let sign decide direction. In Questions 24, 25 and 28 the loop equation settles everything, and in Questions 20 and 30 the potentiometer is just a uniform potential gradient along a wire.

Very Short Answer Questions

Question 12

Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?

Solution. No. At a junction the electrons change direction, and the force that turns them comes from the electric fields of surface charges that accumulate on the wire surfaces near the bend. These fields change the direction of the electrons’ momentum, so their momentum is not conserved. The quantity that is conserved at a junction is charge: the current entering equals the current leaving, which is Kirchhoff’s junction rule.

Question 13

The relaxation time $\tau$ is nearly independent of applied $E$ field whereas it changes significantly with temperature $T$. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of $\rho$ with temperature. Elaborate why?

Solution. The relaxation time is the average time between collisions, and it depends on how fast the electrons are moving. Their speeds are mostly thermal, of order $10^{2}$ m/s (in a metal even more), whereas an applied field adds only a drift of the order of a millimetre per second. The field therefore barely changes the speeds, so $\tau$ is independent of $E$. The drift velocity is then $v_d = (e\tau/m)E$, directly proportional to $E$, and so the current density $j = ne v_d \propto E$. That is Ohm’s law.

Temperature changes the thermal speeds and the amplitude of lattice vibrations strongly, so $\tau$ depends on $T$. Since $\rho = \dfrac{m}{ne^2\tau}$, the resistivity varies with temperature.

Question 14

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R_{\text{unknown}}$ by any other method?

Solution. At balance no current flows through the galvanometer. So the galvanometer’s own resistance and its calibration do not matter, and neither do the emf and internal resistance of the cell that drives the bridge. The result depends only on the ratio of resistances, $R_{\text{unknown}} = R\,\dfrac{l_2}{l_1}$ and so on.

By any other method one would have to measure the currents in every resistor and in the galvanometer accurately, and also know the galvanometer’s internal resistance, and then use Kirchhoff’s laws.

Question 15

What is the advantage of using thick metallic strips to join wires in a potentiometer?

Solution. Thick strips have negligible resistance, so there is no unmeasured potential drop at the joints between the wire segments. Only the straight one-metre lengths of the wire carry the potential gradient, and those are measured with an ordinary metre ruler, so the balance length is read accurately.

Question 16

For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?

Solution. Wires for the home must have low resistivity so that little power is lost as heat, and they must be affordable. Silver conducts best but is far too costly. Copper is next best and is widely used; aluminium has a higher resistivity but is much cheaper and lighter. Mechanical strength and weight can also be mentioned.

Question 17

Why are alloys used for making standard resistance coils?

Solution. A standard resistance must keep its value when the temperature changes slightly as it carries current. Alloys such as manganin and constantan have a very small temperature coefficient of resistance. They also have a high resistivity, so a short length of wire gives a large resistance and the coil stays compact.

Question 18

Power $P$ is to be delivered to a device via transmission cables having resistance $R_c$. If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.

Solution. The cable carries the same current $I$ as the device, so the power wasted in the cables is $P_c = I^2R_c$. The power delivered is $P = VI$, so $I = P/V$ and

$$P_c = \frac{P^2R_c}{V^2}.$$

For fixed $P$ and $R_c$ the loss falls as the square of the voltage, so it is reduced by transmitting at a high voltage (and stepping down near the consumer).

Question 19

AB is a potentiometer wire (Fig 3.4). If the value of $R$ is increased, in which direction will the balance point J shift?

Solution. The figure shows a cell $E$ in series with the rheostat $R$ across the whole wire AB, and a second cell with a galvanometer connecting A to the jockey J. Increasing $R$ lowers the current in the wire. The potential gradient along the wire (volts per unit length) therefore falls. The cell being balanced needs the same potential drop as before, so a longer length of wire is needed to supply it. The balance point moves away from A, towards B.

Question 20

While doing an experiment with potentiometer (Fig 3.5) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased while the jockey was moved towards the end B. (i) Which terminal $+$ or $-$ ve of the cell $E_1$ is connected at X in case (i) and how is $E_1$ related to $E$? (ii) Which terminal of the cell $E_1$ is connected at X in case (ii)?

Solution. The driver cell $E$ has its positive plate on the A side, so the potential of the wire falls steadily from A to B. Take the potential at A as $V_A$ and let the wire have potential gradient $k$, so a point at distance $x$ from A is at $V_A – kx$. The total drop along the wire is $kL = E$, since the driver has no other resistance in its circuit.

The cell $E_1$ sits between X (joined to A) and y (joined through the galvanometer to the jockey J). The galvanometer therefore reads the difference between the potential of y and the potential of J.

(i) Suppose the positive terminal of $E_1$ is at X. Going from A through $E_1$ from its $+$ to its $-$ terminal, the potential drops by $E_1$, so $V_y = V_A – E_1$. The difference across the galvanometer is $$V_J – V_y = (V_A – kx) – (V_A – E_1) = E_1 – kx.$$ This falls as $x$ grows, so the deflection decreases as the jockey moves from A towards B. It never reaches zero, and never changes side, only if $E_1 – kx > 0$ all the way to B, that is $E_1 > kL = E$. So the positive terminal of $E_1$ is at X and $E_1 > E$; this is why no balance point can be found.

(ii) Now put the negative terminal of $E_1$ at X. Going from A through $E_1$ from $-$ to $+$, the potential rises by $E_1$, so $V_y = V_A + E_1$ and $$V_J – V_y = -(E_1 + kx).$$ Its size $E_1 + kx$ grows with $x$, so the deflection is one-sided and increases as the jockey moves towards B, whatever the value of $E_1$. The negative terminal of $E_1$ is at X.

Question 21

A cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$. Plot a graph showing the variation of P.D. across $R$, versus $R$.

Solution. The current is $I = \dfrac{E}{R+r}$, so the potential difference across $R$ is

$$V = IR = \frac{ER}{R+r}.$$

For $R = 0$ the cell is short-circuited and $V = 0$. The slope is $\dfrac{dV}{dR} = \dfrac{Er}{(R+r)^2}$, which is positive everywhere and largest at $R = 0$, where it equals $E/r$. So the curve leaves the origin steeply, rises, and bends over (it is concave down throughout, with no inflection). At $R = r$ it reaches $V = E/2$, and as $R \to \infty$ it approaches the asymptote $V = E$ without reaching it. In words: the graph is a rectangular-hyperbola-like curve rising from the origin and levelling off at $E$.

Printed key error. The key draws an S-shaped curve that starts flat at the origin. That cannot be right: $dV/dR = Er/(R+r)^2$ is greatest at $R=0$, so the true graph starts with its steepest slope and has no flat start and no point of inflection.

Short Answer Questions

Question 22

First a set of $n$ equal resistors of $R$ each are connected in series to a battery of emf $E$ and internal resistance $R$. A current $I$ is observed to flow. Then the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is $n$?

Solution. In series the total resistance is $nR + R$, so $$I = \frac{E}{R + nR} = \frac{E}{R(n+1)}.$$ In parallel the $n$ resistors give $R/n$, so $$10I = \frac{E}{R + R/n} = \frac{nE}{R(n+1)}.$$ Dividing the second equation by the first gives $\dfrac{n}{1} = 10$, so $n = 10$.

Question 23

Let there be $n$ resistors $R_1, \dots, R_n$ with $R_{\max} = \max(R_1,\dots,R_n)$ and $R_{\min} = \min(R_1,\dots,R_n)$. Show that when they are connected in parallel, the resultant resistance $R_p < R_{\min}$ and when they are connected in series, the resultant resistance $R_s > R_{\max}$. Interpret the result physically.

Solution. In parallel, $\dfrac{1}{R_p} = \sum_{i=1}^{n}\dfrac{1}{R_i}$. Multiply by $R_{\min}$: $$\frac{R_{\min}}{R_p} = \sum_{i=1}^{n}\frac{R_{\min}}{R_i}.$$ Each term is at most $1$, and the term for the smallest resistor itself equals exactly $1$. With every other term positive and $n \ge 2$, the sum exceeds $1$, so $R_{\min}/R_p > 1$ and $R_p < R_{\min}$.

In series, $R_s = \sum R_i = R_{\max} + (\text{the other } n-1 \text{ positive resistances}) > R_{\max}$.

Physical meaning. In parallel every extra resistor offers one more path for the charge, so the total current exceeds what the least resistive branch alone would carry and the combined resistance is lower than any single one. In series every extra resistor adds more obstruction along the same path, so the total is greater than any single one.

Question 24

The circuit in Fig 3.6 shows two cells connected in opposition to each other. Cell $E_1$ is of emf 6 V and internal resistance $2\ \Omega$; the cell $E_2$ is of emf 4 V and internal resistance $8\ \Omega$. Find the potential difference between the points A and B.

Solution. The figure shows the two cells in a single loop with A between them and B on the far side of $E_2$. The cells oppose, so the stronger cell $E_1$ drives the current and the net emf is $6 – 4 = 2$ V, against a total resistance of $2 + 8 = 10\ \Omega$: $$I = \frac{6-4}{2+8} = 0.2\ \text{A}.$$ Cell $E_1$ is delivering current, so the potential difference across its terminals is $6 – 0.2\times2 = 5.6$ V. Check with $E_2$: it is being charged, so its terminal potential difference is $4 + 0.2\times8 = 5.6$ V. The two agree. The current flows through B into the positive terminal of $E_2$, so B is at the higher potential.

Question 25

Two cells of same emf $E$ but internal resistance $r_1$ and $r_2$ are connected in series to an external resistor $R$ (Fig 3.7). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero.

Solution. The two cells aid each other in series, so $$I = \frac{2E}{r_1 + r_2 + R}.$$ The terminal voltage of the first cell is $E – Ir_1$. Setting it to zero gives $I = E/r_1$. Then $$\frac{2E}{r_1+r_2+R} = \frac{E}{r_1} \;\Rightarrow\; 2r_1 = r_1 + r_2 + R \;\Rightarrow\; R = r_1 – r_2.$$ This needs $r_1 > r_2$, otherwise no positive $R$ makes the voltage zero.

Question 26

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance $R_A$ to $R_B$.

Solution. For the same material and length, $R = \rho l/A$, so $R \propto 1/A$. The cross-section of A is $\pi(0.5)^2 = 0.25\pi\ \text{mm}^2$. The cross-section of B is the annulus $\pi(1)^2 – \pi(0.5)^2 = 0.75\pi\ \text{mm}^2$. Therefore $$\frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{0.75\pi}{0.25\pi} = 3.$$

Question 27

Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

Solution. Apply Kirchhoff’s loop rule to any loop: $\sum V_k = \sum I_j R_j$, and the junction rule $\sum I = 0$. Replace every voltage by $nV_k$ and every resistance by $nR_j$. The loop equations become $n\sum V_k = \sum I_j’ (nR_j)$, and dividing by $n$ gives exactly the original equations. The junction equations do not contain $V$ or $R$ at all. The same set of equations therefore has the same solution, so $I_j’ = I_j$.

Another way to see it: any branch with resistance $R$ carries a current $V_{\text{eff}}/(R_{\text{eff}} + R)$, where $V_{\text{eff}}$ is proportional to the voltages and $R_{\text{eff}}$ is built from the resistances. Scaling both by $n$ cancels $n$ in the ratio.

For the circuit of Example 3.7 the same argument applies line by line: each loop equation of that example is multiplied by $n$ on both sides, so the currents found there come out identical for any $n$.

Long Answer Questions

Question 28

Two cells of voltage 10 V and 2 V and internal resistances $10\ \Omega$ and $5\ \Omega$ respectively, are connected in parallel with the positive end of 10 V battery connected to negative pole of 2 V battery (Fig 3.8). Find the effective voltage and effective resistance of the combination.

Solution. Treat the cells as emfs $E_1 = 10$ V with $r_1 = 10\ \Omega$ and $E_2 = -2$ V with $r_2 = 5\ \Omega$, the negative sign because the 2 V cell is connected the opposite way round. For cells in parallel $$V_{\text{eff}} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{10/10 – 2/5}{1/10 + 1/5} = \frac{0.6}{0.3} = 2\ \text{V},$$ $$R_{\text{eff}} = \frac{r_1r_2}{r_1+r_2} = \frac{10\times5}{15} = \frac{10}{3}\ \Omega.$$

Question 29

A room has AC run for 5 hours a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? $[\rho_{\text{Cu}} = 1.7\times10^{-8}\ \Omega\,\text{m},\ \rho_{\text{Al}} = 2.7\times10^{-8}\ \Omega\,\text{m}]$

Solution. Ten units is 10 kWh used over 5 hours, so the power is $P = 2$ kW. The current is $I = \dfrac{2000}{220} = 9.09$ A.

Copper: $R = \dfrac{\rho l}{A} = \dfrac{1.7\times10^{-8}\times10}{\pi\times(10^{-3})^2} = 0.0541\ \Omega$. The heating loss is $I^2R = (9.09)^2\times0.0541 = 4.47$ W, which is $\dfrac{4.47}{2000} = 0.22\%$ of the power.

Aluminium: $R = \dfrac{2.7\times10^{-8}\times10}{\pi\times10^{-6}} = 0.0859\ \Omega$, so the loss is $7.10$ W, or $0.36\%$. It is larger in the ratio $2.7/1.7 \approx 1.6$.

(The key rounds $I$ to 9 A and the ratio to 1.6 and so quotes about 4 W for copper and 6.4 W for aluminium. That is rounding, not an error.)

Question 30

In an experiment with a potentiometer, $V_B = 10$ V. $R$ is adjusted to be $50\ \Omega$ (Fig 3.9). A student wanting to measure voltage $E_1$ of a battery (approx. 8 V) finds no null point possible. He then diminishes $R$ to $10\ \Omega$ and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Solution. The potentiometer wire, of resistance $R’$ and length 400 cm (four segments of 1 m), is in series with the rheostat $R$ and the 10 V battery. The voltage across the whole wire is $$V_w = \frac{10R’}{R + R’}.$$

First case, $R = 50\ \Omega$: no null point exists, so even the whole wire does not reach 8 V: $$\frac{10R’}{50+R’} < 8 \;\Rightarrow\; 10R’ < 400 + 8R’ \;\Rightarrow\; R’ < 200\ \Omega.$$

Second case, $R = 10\ \Omega$: the null point is on the 4th segment, so the drop over the whole wire exceeds 8 V: $$\frac{10R’}{10+R’} > 8 \;\Rightarrow\; 2R’ > 80 \;\Rightarrow\; R’ > 40\ \Omega.$$ The null point lies in the last segment, so it is also needed that the drop over the first 300 cm is below 8 V: $\dfrac{3}{4}\cdot\dfrac{10R’}{10+R’} = \dfrac{7.5R’}{10+R’} < 8$, which means $7.5R’ < 80 + 8R’$. That holds for every positive $R’$, so it adds no further restriction.

Hence $40\ \Omega < R’ < 200\ \Omega$.

The potential gradient in the second case is $$\varphi = \frac{10R’}{(10+R’)\times4\ \text{m}}.$$ At $R’ = 40\ \Omega$ it is $\dfrac{8}{4} = 2$ V/m, and as $R’\to200\ \Omega$ it rises to $\dfrac{9.52}{4} = 2.38$ V/m. So $2\ \text{V/m} < \varphi < 2.38\ \text{V/m}$, that is $0.020$ to $0.024$ V/cm.

Printed key error. The key reduces the 300 cm condition to $R’ > 160\ \Omega$ through a sign slip and so reports $160\ \Omega < R’ < 200\ \Omega$. In fact $7.5R’ < 80 + 8R’$ is true for every positive $R’$, so the lower bound comes only from the whole-wire condition and is $40\ \Omega$. The key’s gradient range $2 < \varphi < 2\tfrac23$ V/m is true but its upper bound is never reached; the correct upper bound is about $2.38$ V/m.

Question 31

(a) Consider circuit in Fig 3.10. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity? (b) Electrons give up energy at the rate of $RI^2$ per second to the thermal energy. What time scale would one associate with energy in problem (a)? $n$ = no of electron/volume = $10^{29}/\text{m}^3$, length of circuit = 10 cm, cross-section = $A = (1\ \text{mm})^2$.

Solution. The circuit is a 6 V cell driving a $6\ \Omega$ resistor, so $I = 6/6 = 1$ A.

(a) The drift velocity is $$v_d = \frac{I}{neA} = \frac{1}{10^{29}\times1.6\times10^{-19}\times10^{-6}} = 6.25\times10^{-5}\ \text{m/s}.$$ The number of electrons is $N = nAl = 10^{29}\times10^{-6}\times0.1 = 10^{22}$. The energy they absorb is their kinetic energy, $$KE = N\cdot\tfrac12 m v_d^2 = 10^{22}\times\tfrac12\times9.1\times10^{-31}\times(6.25\times10^{-5})^2 \approx 1.78\times10^{-17}\ \text{J},$$ about $2\times10^{-17}$ J.

(b) Electrons give up energy at the rate $RI^2 = 6\times1 = 6$ W, so the time scale is $$t = \frac{1.78\times10^{-17}}{6} \approx 3\times10^{-18}\ \text{s}.$$ This is far shorter than anything measurable, which shows that the drift kinetic energy is negligible and the current is established essentially instantly.

Printed key error. Minor: the key writes the time scale as about $10^{-17}$ s, but its own numbers give $3.3\times10^{-18}$ s, so $10^{-17}$ s is an order-of-magnitude slip.