NCERT Exemplar Solutions Class 12 Physics Alternating Current — Very Short, Short and Long Answer Questions Q14–31

NCERT Exemplar Problems, Class 12 Physics — Alternating Current, Very Short Answer Questions 14 to 20, Short Answer Questions 21 to 26 and Long Answer Questions 27 to 31. All 18 questions solved. The book numbers them 7.14 to 7.31.

Three ideas carry this page. Reactance depends on frequency: $X_L=\omega L$ grows and $X_C=1/\omega C$ shrinks as $\omega$ rises, so at high frequency an inductor behaves like an open circuit and a capacitor like a short. Only the resistive part dissipates: the average power is $VI\cos\phi$, and $\cos\phi=R/Z$ is the whole story behind Questions 17, 27 and 30. The LCR circuit is a damped, driven oscillator: the energy bookkeeping in Questions 14, 30 and 31 is the mechanical spring-block analogy with $L\leftrightarrow m$ and $1/C\leftrightarrow k$.

Key insight. Always find $Z$ and $\phi$ first, then everything else is a ratio. In Question 27 the power factor $\cos\phi$ must be kept in $P=\dfrac{V^2}{Z}\cos\phi$, and in Question 28 the entire point of a transformer is that $I^2R$ loss falls as the square of the transmission voltage.

Very Short Answer Questions

Question 14

If a $LC$ circuit is considered analogous to a harmonically oscillating spring block system, which energy of the $LC$ circuit would be analogous to potential energy and which one analogous to kinetic energy?

Solution. Compare the two equations of motion. The block obeys $m\ddot x+kx=0$ and the circuit obeys $L\ddot q+\dfrac{q}{C}=0$, so the correspondence is $m\leftrightarrow L$, $k\leftrightarrow 1/C$, $x\leftrightarrow q$ and velocity $\leftrightarrow$ current $i=\dot q$. Then the kinetic energy $\tfrac12 mv^2$ matches the magnetic energy $\tfrac12 Li^2$ stored in the inductor, and the potential energy $\tfrac12 kx^2$ matches the electric energy $\dfrac{q^2}{2C}$ stored in the capacitor.

Question 15

Draw the effective equivalent circuit of the circuit shown in Fig 7.1, at very high frequencies and find the effective impedance.

Solution. The figure has a source across two branches. The upper branch is $R_1$, then $C_1$, then $L_1$, with a second capacitor $C_2$ hanging down from the point between $C_1$ and $L_1$ to the point between $R_2$ and $R_3$ in the lower branch. The lower branch is $L_2$, $R_2$ and $R_3$ in series.

At very high frequency $\omega\to\infty$, a capacitor has $X_C=1/\omega C\to0$ and acts as a plain wire, while an inductor has $X_L=\omega L\to\infty$ and acts as a break. So $C_1$ and $C_2$ become wires and $L_1$ and $L_2$ become open circuits. Current cannot pass through $L_1$ or $L_2$, so the only closed path is the source, $R_1$, $C_1$ (wire), $C_2$ (wire), $R_3$ and back to the source. The resistor $R_2$ is in series with the open $L_2$, so it carries no current and simply dangles. The equivalent circuit is $R_1$ in series with $R_3$.

Question 16

Study the circuits (a) and (b) shown in Fig 7.2 and answer the following questions. (a) Under what conditions would the rms currents in the two circuits be the same? (b) Can the rms current in circuit (b) be larger than that in (a)?

Solution. Circuit (a) is a resistor $R$ alone across the source. Circuit (b) is $R$, $C$ and $L$ in series across the same kind of source.

(a) With the same rms source voltage and the same $R$, the currents are $I_a=\dfrac{V}{R}$ and $I_b=\dfrac{V}{\sqrt{R^2+(X_L-X_C)^2}}$. They are equal only when $X_L=X_C$, that is at resonance, $\omega=1/\sqrt{LC}$.

(b) No. The impedance of circuit (b) is $Z=\sqrt{R^2+(X_L-X_C)^2}\ge R$, so $I_b=V/Z\le V/R=I_a$. The reactive part can only add opposition; at best it cancels completely.

Question 17

Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?

Solution. The instantaneous power is $p=vi$. When the voltage and current have opposite signs, $p$ is negative; this happens in part of every cycle whenever there is a phase difference, because the inductor or capacitor hands stored energy back to the source. So the instantaneous power can be negative.

The average power is $\bar P=VI\cos\phi$, with rms values $V$ and $I$. In a passive circuit the phase difference satisfies $|\phi|\le\pi/2$, so $\cos\phi\ge0$ and $\bar P\ge0$. The energy returned during part of the cycle never exceeds what was supplied, and the resistor dissipates a positive amount on average. The average power cannot be negative.

Question 18

In series LCR circuit, the plot of $I_{\max}$ vs $\omega$ is shown in Fig 7.3. Find the bandwidth and mark in the figure.

Solution. The curve peaks at $\omega_0=1.0$ rad/s where the current amplitude is $I_{\max}=1.0$ A. The half-power points are where the current amplitude falls to $I_{\max}/\sqrt2\approx0.7$ A.

To mark it, draw a horizontal line across the graph at $I_m\approx0.7$ A. It meets the resonance curve at two points, one on each side of the peak, at about $\omega_1\approx0.8$ rad/s and $\omega_2\approx1.2$ rad/s. Mark these on the $\omega$-axis; the bandwidth is the interval between them:

$$\Delta\omega=\omega_2-\omega_1\approx1.2-0.8=0.4\ \text{rad/s}.$$

Question 19

The alternating current in a circuit is described by the graph shown in Fig 7.4. Show rms current in this graph.

Solution. Read the graph over one full period, which is $2T$. The current is $+1$ A from $0$ to $T/2$, then $-2$ A from $T/2$ to $3T/2$, then $+1$ A from $3T/2$ to $2T$. The rms current is the square root of the mean of $I^2$ over a period, so the signs do not matter:

$$\langle I^2\rangle=\frac{1^2\cdot\frac T2+(-2)^2\cdot T+1^2\cdot\frac T2}{2T}=\frac{\frac T2+4T+\frac T2}{2T}=\frac{5T}{2T}=2.5\ \text{A}^2.$$

$$I_{\text{rms}}=\sqrt{2.5}\approx1.6\ \text{A}.$$

On the graph, draw a horizontal line at $I=1.6$ A (between the $1$ A and $2$ A marks, a little above the $1.5$ A level) across the whole period.

Question 20

How does the sign of the phase angle $\phi$, by which the supply voltage leads the current in an $LCR$ series circuit, change as the supply frequency is gradually increased from very low to very high values.

Solution. The phase angle satisfies $\tan\phi=\dfrac{X_L-X_C}{R}$. At very low frequency $X_C=1/\omega C$ is huge and $X_L=\omega L$ is tiny, so $X_L-X_C<0$ and $\phi$ is negative: the circuit is capacitive and the voltage lags the current. As $\omega$ rises, $X_C$ falls and $X_L$ rises, so $\phi$ increases steadily. At the resonant frequency $\omega_0=1/\sqrt{LC}$ the reactances are equal, $\tan\phi=0$ and $\phi=0$. Above $\omega_0$, $X_L>X_C$, so $\phi$ is positive and the circuit is inductive; at very high frequency $\phi\to+\pi/2$.

Short Answer Questions

Question 21

A device ‘X’ is connected to an a.c source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5. (a) Which curve shows power consumption over a full cycle? (b) What is the average power consumption over a cycle? (c) Identify the device ‘X’.

Solution. The figure shows three curves A, B and C against time. Curve A has larger amplitude and goes through two complete oscillations while B and C go through one. B and C have the same period and are a quarter-cycle (90°) apart.

(a) Power is $p=vi$, the product of two sinusoids of frequency $f$, so it oscillates at $2f$. Curve A has twice the frequency of the others, so A is the power curve.

(b) Since $v$ and $i$ are $90^\circ$ apart, $p=v_mi_m\sin\omega t\cos\omega t=\tfrac12v_mi_m\sin2\omega t$, which spends equal time positive and negative. Its average is zero.

(c) A device that draws current $90^\circ$ out of phase with the voltage and consumes no average power is purely reactive: a pure inductor, a pure capacitor, or an LC combination, with no resistance.

Question 22

Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?

Solution. The usual definition of the ampere uses the magnetic force between two parallel current-carrying wires. For ac that cannot be used directly: the average of an alternating current over a cycle is zero, and the force between two wires carrying ac also averages to zero. We need an effect that does not depend on the direction of the current, and that is the heating effect. So a current of 1 A rms is that alternating current which produces the same Joule heat in a resistor in a given time as a steady direct current of 1 A.

Question 23

A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.

Solution. The angular frequency is $\omega=2\pi\times50=100\pi$ rad/s, so the inductive reactance is

$$X_L=\omega L=100\pi\times0.01=\pi\approx3.14\ \Omega.$$

The impedance is

$$Z=\sqrt{R^2+X_L^2}=\sqrt{1+\pi^2}\approx3.3\ \Omega.$$

The current lags the voltage by $\phi$ with

$$\tan\phi=\frac{X_L}{R}=\pi,\qquad \phi=\tan^{-1}\pi\approx72^\circ\approx1.26\ \text{rad}.$$

The time lag between the maximum of the voltage and the maximum of the current is

$$\Delta t=\frac{\phi}{\omega}=\frac{1.26}{100\pi}\approx4.0\times10^{-3}\ \text{s}\approx\frac1{250}\ \text{s}.$$

Question 24

A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.

Solution. The load draws 60 W at 0.54 A, so the secondary voltage is

$$V_s=\frac{60}{0.54}\approx110\ \text{V}.$$

For an ideal transformer the input power equals the output power. The primary draws the 220 V line, so

$$I_p=\frac{P}{V_p}=\frac{60}{220}\approx0.27\ \text{A}.$$

The secondary voltage ($110$ V) is lower than the primary voltage ($220$ V), and the secondary current is larger than the primary current, so this is a step-down transformer with a turns ratio $N_p:N_s=2:1$.

Question 25

Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

Solution. The current into a capacitor is the rate at which its charge changes, $i=\dfrac{dq}{dt}=C\dfrac{dv}{dt}$. When the frequency of the applied voltage rises, the voltage changes faster, so $dv/dt$ is larger and more charge must flow on and off the plates each second. For the same voltage amplitude the current is therefore larger, which means the opposition to the current is smaller. Quantitatively $X_C=\dfrac1{\omega C}$, which falls as $\omega$ rises. At very low frequency the capacitor has no time to pass charge and blocks the current, and in the limit of dc it is an open circuit.

Question 26

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

Solution. When the current through an inductor changes, a back emf $\mathcal E=-L\dfrac{di}{dt}$ is induced which, by Lenz’s law, opposes the change. The higher the frequency, the faster the current tries to change, so $di/dt$ is larger and the back emf is larger. The source must work harder against this opposing emf, so for the same voltage amplitude the current is smaller. That is the same as saying the reactance rises: $X_L=\omega L$, increasing with $\omega$. At dc there is no change, no back emf and no reactance.

Long Answer Questions

Question 27

An electrical device draws 2 kW power from AC mains (voltage 223 V (rms) $=\sqrt{50{,}000}$ V). The current differs (lags) in phase by $\phi\left(\tan\phi=\dfrac{-3}{4}\right)$ as compared to voltage. Find (i) $R$, (ii) $X_C-X_L$, and (iii) $I_M$. Another device has twice the values for $R$, $X_C$ and $X_L$. How are the answers affected?

Solution. The key quantity is the power factor. With $\tan\phi=-\tfrac34$ we have $|\cos\phi|=\dfrac45$. The average power is $P=V_{\text{rms}}I_{\text{rms}}\cos\phi$ and $I_{\text{rms}}=V_{\text{rms}}/Z$, so

$$P=\frac{V_{\text{rms}}^2}{Z}\cos\phi\;\Rightarrow\; Z=\frac{V_{\text{rms}}^2\cos\phi}{P}=\frac{50000\times0.8}{2000}=20\ \Omega.$$

(i) Since $\cos\phi=R/Z$, the resistance is $R=Z\cos\phi=20\times0.8=16\ \Omega$.

(ii) Following the question’s convention, $\tan\phi=\dfrac{X_C-X_L}{R}=-\dfrac34$, so $$X_C-X_L=R\tan\phi=16\times\left(-\tfrac34\right)=-12\ \Omega.$$ The negative sign means the device is inductive ($X_L>X_C$), which is why the current lags the voltage.

(iii) The rms current is $I_{\text{rms}}=\dfrac{V_{\text{rms}}}{Z}=\dfrac{\sqrt{50000}}{20}=\dfrac{223.6}{20}\approx11.2$ A, so the peak current is

$$I_M=\sqrt2\,I_{\text{rms}}\approx1.414\times11.18\approx15.8\ \text{A}.$$

Check: $I_{\text{rms}}^2R=125\times16=2000$ W, as required.

Doubled device. If $R$, $X_C$ and $X_L$ are all doubled, then $\tan\phi=\dfrac{X_L-X_C}{R}$ is unchanged, so the phase angle is the same. The impedance doubles, $Z=40\ \Omega$. At the same supply voltage the current halves, so $I_M\approx7.9$ A. The power $P=I^2R$ becomes $\left(\tfrac12I\right)^2\times2R=\tfrac12I^2R$, half of before, so $P=1$ kW.

Printed key error. The key uses $P=V^2/Z$, which drops the power factor, and so gets $Z=25\ \Omega$, $R=20\ \Omega$, $X_C-X_L=-15\ \Omega$, $I\approx9$ A and $I_M=12.6$ A. Those values do not even reproduce the given power: $I^2R=(223.6/25)^2\times20=1600$ W, not 2000 W. The correct route keeps $\cos\phi$ as above. The key’s conclusions for the doubled device (same $\tan\phi$, impedance doubles, current halves, power halves) are right.

Question 28

1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if (i) power is transmitted at 220 V. Comment on the feasibility of doing this. (ii) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V. ($\rho_{Cu}=1.7\times10^{-8}$ SI unit)

Solution. A pair of wires over 10 km means a total conductor length $l=2\times10\ \text{km}=2\times10^4$ m, with cross-section $A=\pi(0.5\times10^{-2})^2=7.85\times10^{-5}\ \text{m}^2$. The line resistance is

$$R=\frac{\rho l}{A}=\frac{1.7\times10^{-8}\times2\times10^4}{7.85\times10^{-5}}\approx4.3\ \Omega.$$

(i) At 220 V the current for 1 MW is $I=\dfrac{10^6}{220}\approx4.5\times10^3$ A. The loss in the line is $I^2R\approx(4.5\times10^3)^2\times4.3\approx8.9\times10^7$ W, which is about 89 times the power to be delivered. That is absurd: the lines would dissipate far more than the station could supply, so transmission at 220 V is not feasible.

(ii) At 11000 V the line current is $I=\dfrac{10^6}{11000}\approx90.9$ A. The loss is $I^2R\approx(90.9)^2\times4.3\approx3.6\times10^4$ W, so the fraction lost is

$$\frac{3.6\times10^4}{10^6}\approx0.036=3.6\%.$$

Raising the voltage fifty-fold cut the current fifty-fold and the loss by a factor of $50^2=2500$. (The key rounds $R$ to $4\ \Omega$ and quotes $3.3\%$; that is rounding, not an error.)

Question 29

Consider the LCR circuit shown in Fig 7.6. Find the net current $i$ and the phase of $i$. Show that $i=\dfrac{v}{Z}$. Find the impedance $Z$ for this circuit.

Solution. In Fig 7.6 the resistor $R$ is in parallel with a series combination of $C$ and $L$, and both branches are connected across the source $v=v_m\sin\omega t$. The net current $i=i_1+i_2$ splits between the two branches.

Resistor branch. The voltage across $R$ is the source voltage, so $$i_1=\frac{v_m}{R}\sin\omega t,$$ in phase with the voltage.

LC branch. The charge on the capacitor obeys $\dfrac{q_2}{C}+L\ddot q_2=v_m\sin\omega t$. Try $q_2=q_m\sin\omega t$. Then $q_m\left(\dfrac1C-L\omega^2\right)=v_m$, so $$q_2=\frac{v_m}{\frac1C-L\omega^2}\sin\omega t.$$ The current is $i_2=\dot q_2=\dfrac{\omega v_m}{\frac1C-L\omega^2}\cos\omega t$, and dividing top and bottom by $\omega$, $$i_2=\frac{v_m}{X_C-X_L}\cos\omega t,\qquad X_C=\frac1{\omega C},\ X_L=\omega L.$$ This branch is $90^\circ$ out of phase with $i_1$.

Net current. Add the two perpendicular components: $$i=i_1+i_2=\frac{v_m}{R}\sin\omega t+\frac{v_m}{X_C-X_L}\cos\omega t=\frac{v_m}{Z}\sin(\omega t+\phi),$$ with amplitude and phase fixed by $$\frac1Z=\sqrt{\frac1{R^2}+\frac1{(X_L-X_C)^2}},\qquad \tan\phi=\frac{R}{X_C-X_L}.$$ So the net current is $i=\dfrac{v}{Z}$ in the sense of amplitudes, $i_m=v_m/Z$, with $Z$ given above. The current leads the voltage when the LC branch is capacitive ($X_C>X_L$) and lags it when the branch is inductive ($X_L>X_C$).

Printed key error. The key writes $i_2=\dfrac{v_m}{L\omega-1/C\omega}\cos\omega t$ and $\phi=\tan^{-1}\dfrac{R}{X_L-X_C}$. That is inconsistent with its own charge amplitude $q_m=v_m/(1/C-L\omega^2)$, which puts $X_C-X_L$ in the denominator. The sign of $\phi$ in the key is therefore reversed; $Z$ is unaffected.

Question 30

For an LCR circuit driven at frequency $\omega$, the equation reads $L\dfrac{di}{dt}+Ri+\dfrac qC=v_i=v_m\sin\omega t$. (i) Multiply the equation by $i$ and simplify where possible. (ii) Interpret each term physically. (iii) Cast the equation in the form of a conservation of energy statement. (iv) Integrate the equation over one cycle to find that the phase difference between $v$ and $i$ must be acute.

Solution.

(i) Multiplying by $i$: $$Li\frac{di}{dt}+Ri^2+\frac qCi=vi.$$ The first and third terms are exact derivatives, because $i=\dfrac{dq}{dt}$: $Li\dfrac{di}{dt}=\dfrac{d}{dt}\left(\tfrac12Li^2\right)$ and $\dfrac qCi=\dfrac{d}{dt}\left(\dfrac{q^2}{2C}\right)$.

(ii) The terms mean: $\dfrac{d}{dt}\left(\tfrac12Li^2\right)$ is the rate of change of the energy stored in the magnetic field of the inductor; $Ri^2$ is the rate at which energy is dissipated as Joule heat in the resistor; $\dfrac{d}{dt}\left(\dfrac{q^2}{2C}\right)$ is the rate of change of the electric energy stored in the capacitor; and $vi$ is the power supplied by the source.

(iii) Combining the exact derivatives gives $$\frac{d}{dt}\left(\tfrac12Li^2+\frac{q^2}{2C}\right)+Ri^2=vi.$$ Power supplied by the source equals the rate of increase of stored energy plus the rate of dissipation: energy is conserved.

(iv) Integrate over one period $T$ in the steady state. The stored energy $\tfrac12Li^2+\dfrac{q^2}{2C}$ has the same value at the end as at the start, so the first term integrates to zero: $$\int_0^Tvi\,dt=\int_0^TRi^2\,dt>0.$$ With $v=v_m\sin\omega t$ and $i=i_m\sin(\omega t+\phi)$, the left side is $\dfrac{v_mi_mT}{2}\cos\phi$. Since it must be positive, $\cos\phi>0$, so $|\phi|<\pi/2$: the phase difference between $v$ and $i$ is acute.

(The key’s “$\tfrac12i^2$” for the magnetic energy omits the factor $L$; it is a misprint and does not change the argument.)

Question 31

In the LCR circuit shown in Fig 7.7, the ac driving voltage is $v=v_m\sin\omega t$. (i) Write down the equation of motion for $q(t)$. (ii) At $t=t_0$, the voltage source stops and $R$ is short circuited. Now write down how much energy is stored in each of $L$ and $C$. (iii) Describe subsequent motion of charges.

Solution. The figure shows the source, $R$ (with a switch across it for short-circuiting), $L$ and $C$ all in one series loop.

(i) The loop equation with $i=\dot q$ is $$L\ddot q+R\dot q+\frac qC=v_m\sin\omega t.$$ In the steady state the charge oscillates at the driving frequency, $q=-q_m\cos(\omega t+\phi)$, so that $i=\dot q=i_m\sin(\omega t+\phi)$ with $$i_m=\omega q_m=\frac{v_m}{\sqrt{R^2+(X_C-X_L)^2}},\qquad \tan\phi=\frac{X_C-X_L}{R}.$$

(ii) At $t=t_0$ the source is removed and $R$ is bypassed, leaving a bare $LC$ circuit. The energy in the inductor is $$U_L=\tfrac12Li_m^2\sin^2(\omega t_0+\phi),$$ and in the capacitor $$U_C=\frac{q^2}{2C}=\frac{i_m^2}{2C\omega^2}\cos^2(\omega t_0+\phi).$$

(iii) With no source and no resistance the charges execute undamped $LC$ oscillations at the natural frequency $\omega_0=\dfrac1{\sqrt{LC}}$, whatever the original driving frequency $\omega$. Energy passes back and forth between the capacitor (electric) and the inductor (magnetic), exactly as in a spring-block system, and the total $U_L+U_C$ stays constant at its value at $t_0$.