NCERT Exemplar Solutions Class 12 Physics Alternating Current — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Alternating Current, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 7.1 to 7.13; here they are Questions 1 to 13. Questions 1 to 7 are MCQ I (exactly one option is correct) and Questions 8 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• rms values. A meter or a bulb responds to $\sqrt{\langle v^2\rangle}$, so “$220$ V mains” is an rms figure and the peak is $\sqrt2$ times larger.
• Resonance. In a series LCR circuit $f_0 = 1/(2\pi\sqrt{LC})$, and the sharpness of tuning is the quality factor $Q = \frac1R\sqrt{L/C}$.
• Power factor. The average power is $VI\cos\phi$ with $\cos\phi = R/Z \ge 0$, so a source never takes back energy on average.
• Transmission. Stepping the voltage up lowers the current for the same power, and the loss $I^2R$ falls with the square of the current.

Key insight. Almost every option here is settled by asking what a reactance does to the current: it adds a $90^\circ$ phase shift, stores energy and returns it, and so never dissipates power by itself.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

If the rms current in a $50$ Hz ac circuit is $5$ A, the value of the current $1/300$ seconds after its value becomes zero is

Solution. The peak current is $I_0 = \sqrt2\,I_{rms} = 5\sqrt2$ A. Measure time from a zero, so $i = I_0\sin\omega t$ with $\omega = 2\pi \times 50 = 100\pi$ rad/s. At $t = 1/300$ s the phase is

$$\omega t = \frac{100\pi}{300} = \frac{\pi}{3},$$

so

$$i = 5\sqrt2\,\sin\frac{\pi}{3} = 5\sqrt2\times\frac{\sqrt3}{2} = 5\sqrt{\tfrac32}\ \text{A} \approx 6.1\ \text{A}.$$

The common slip is to give the rms value $5$ A or the peak $5\sqrt2$ A, which are not the instantaneous value at this moment.

Question 2

An alternating current generator has an internal resistance $R_g$ and an internal reactance $X_g$. It is used to supply power to a passive load consisting of a resistance $R_g$ and a reactance $X_L$. For maximum power to be delivered from the generator to the load, the value of $X_L$ is equal to

Solution. The generator and the load are in series, so the same current $I$ flows through both. The total resistance is $R_g + R_g = 2R_g$ and the total reactance is $X_g + X_L$. The current is

$$I = \frac{V}{\sqrt{(2R_g)^2 + (X_g + X_L)^2}},$$

and the power reaching the load is $I^2R_g$. For a fixed $R_g$ this is largest when $I$ is largest, which happens when the reactances cancel, $X_g + X_L = 0$. Hence $X_L = -X_g$: the load must have a reactance of the opposite kind to the generator’s, for instance a capacitive load for an inductive generator. This is the ac form of maximum power transfer, where the load impedance is the complex conjugate of the source impedance.

Question 3

When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of $220$ V. This means

Solution. The mains voltage is $v = v_0\sin\omega t$, which reverses $50$ times a second, far too fast for a pointer to follow. A meter therefore responds to a time average. The plain average of $v$ is zero, which is useless, so the meter responds to $\langle v^2\rangle$ and is marked to display its square root, the rms value. For a sinusoid this is $v_0/\sqrt2$, so $220$ V rms means a peak of $220\sqrt2 \approx 311$ V. That rules out (a), since the supply is genuinely alternating, and (b), since $220$ V is not the peak. A mechanical fault (d) is not needed to explain a steady reading.

Question 4

To reduce the resonant frequency in an LCR series circuit with a generator

Solution. The resonant frequency belongs to the circuit, not to the generator:

$$f_0 = \frac{1}{2\pi\sqrt{LC}}.$$

Lowering $f_0$ means raising $L$ or $C$. Changing the generator frequency (a) only moves the driving frequency away from $f_0$; it leaves $f_0$ itself untouched. A second capacitor in parallel adds capacitances, so $C$ increases and $f_0$ falls: (b). Removing the iron core (c) lowers $L$, and removing the dielectric (d) lowers $C$, since a dielectric raises capacitance by a factor $K$. Both of these raise $f_0$.

Question 5

Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?

Solution. Good tuning means a sharp resonance, so that the circuit picks out one station and rejects its neighbours. The sharpness is measured by the quality factor

$$Q = \frac{1}{R}\sqrt{\frac{L}{C}},$$

and the larger $Q$ is, the narrower the bandwidth. Evaluating each option:

Option (c) has both the smallest resistance and the largest $L/C$, so it wins comfortably.

Question 6

An inductor of reactance $1\ \Omega$ and a resistor of $2\ \Omega$ are connected in series to the terminals of a $6$ V (rms) a.c. source. The power dissipated in the circuit is

Solution. The impedance of the series pair is

$$Z = \sqrt{R^2 + X_L^2} = \sqrt{4 + 1} = \sqrt5\ \Omega,$$

so the rms current is $I = 6/\sqrt5$ A. Only the resistor dissipates power:

$$P = I^2R = \frac{36}{5}\times 2 = 14.4\ \text{W}.$$

The trap is $V^2/R = 18$ W, which ignores the inductor and so overstates the current, and $V^2/Z = 7.2$ W, which leaves out the power factor $\cos\phi = R/Z$. Using $P = VI\cos\phi = 6\times\frac{6}{\sqrt5}\times\frac{2}{\sqrt5} = 14.4$ W gives the same value.

Question 7

The output of a step-down transformer is measured to be $24$ V when connected to a $12$ watt light bulb. The value of the peak current is

Solution. The $24$ V is an rms reading, and a bulb is a resistor, so $\cos\phi = 1$ and

$$I_{rms} = \frac{P}{V_{rms}} = \frac{12}{24} = 0.5\ \text{A}.$$

The peak value is

$$I_0 = \sqrt2\,I_{rms} = \frac{\sqrt2}{2} = \frac{1}{\sqrt2}\ \text{A} \approx 0.71\ \text{A}.$$

Option (b) is what you get by dividing by $\sqrt2$ the wrong way round, and (c) and (d) come from using $V/P$ in place of $P/V$.

Question 8

As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?

Solution. A current that rises and then falls with frequency has a peak, and a peak in current is resonance. Resonance needs the reactance $X_L – X_C$ to pass through zero as $\omega$ increases. At low $\omega$ the capacitor dominates ($X_C = 1/\omega C$ is large) and the current is small; at high $\omega$ the inductor dominates ($X_L = \omega L$ is large) and the current is small again. In between, the two cancel and the current is a maximum. That needs both an inductor and a capacitor, with or without a resistor, so (a) and (d). Resistor with inductor (b) gives a current that only falls with $\omega$, and resistor with capacitor (c) gives a current that only rises towards a limit.

Question 9

In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

Solution. For a series circuit $I = V/Z$, so the current rises with frequency only if the impedance falls. A resistor alone (a) has $Z = R$, independent of frequency. An inductor has $X_L = \omega L$, which grows with $\omega$, so (b) has a current that falls. A capacitor has $X_C = 1/\omega C$, which falls as $\omega$ increases, so the impedance $\sqrt{R^2 + X_C^2}$ of (c) falls and the current rises; for a capacitor alone (d), $I = V\omega C$ rises in direct proportion to $\omega$.

Question 10

Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?

Solution. The power sent is $P = VI$, so for a fixed $P$ a higher $V$ means a lower $I$: (a). The loss in the line is $I^2R$, which falls with the square of the current: (b). Statement (d) is the practical reason for using alternating voltage: a transformer steps the voltage up at the power station and down again near the consumer cheaply and with little loss, which is not possible for dc.

Statement (c) is the tempting one. A thinner wire has a larger resistance $R$, which raises the loss $I^2R$ and undoes the saving. The point of high voltage is to cut the loss, not to save on copper by thinning the line, so (c) is not a correct statement of why the transmission is done.

Question 11

For an LCR circuit, the power transferred from the driving source to the driven oscillator is $P = I^2Z\cos\phi$.

Solution. In a series LCR circuit $Z\cos\phi = R$, so

$$P = I^2Z\cos\phi = I^2R, \qquad \cos\phi = \frac{R}{Z} \ge 0.$$

• (a) Since $R \ge 0$ and $Z > 0$, the power factor is never negative, and so $P \ge 0$. True.
• (b) If $R = 0$ (an ideal LC circuit), then $\cos\phi = 0$ and $P = 0$: the source supplies nothing on average, and the energy just swaps between $L$ and $C$. True.
• (c) The only dissipative element is the resistor, which always takes energy; $P < 0$ would need a negative resistance. On average the source cannot draw energy back out. True.
• (d) This is the opposite of (c), so it is false.

Question 12

When an AC voltage of $220$ V is applied to the capacitor $C$

Solution. The $220$ V is the rms value, so the maximum voltage across the plates is $220\sqrt2 \approx 311$ V, which makes (a) false. For $v = v_0\sin\omega t$, the charge is $q = Cv = Cv_0\sin\omega t$, in phase with the voltage: (c) is true. The current $i = dq/dt = \omega Cv_0\cos\omega t$ leads the voltage by $90^\circ$, so (b) is false. Because the phase difference is $\phi = 90^\circ$, $\cos\phi = 0$ and the average power $VI\cos\phi$ is zero: the capacitor takes energy in one quarter-cycle and returns it in the next. So (d) is true.

Question 13

The line that draws power supply to your house from street has

Solution. The supply is alternating, and a sinusoid has an average of zero over a full cycle. So the average current is zero (a), and the average voltage is also zero, not $220$ V (b); the figure of $220$ V is the rms value. Option (c) would mean a purely reactive load, which draws no net power and is not what a house presents. A real household load is a mixture of resistive heaters and lamps with inductive motors and capacitive electronics, so the current may lag or lead the voltage by some angle $\phi$ with $|\phi| < \pi/2$, so that real power is drawn: (d).

Common mistakes

• Quoting the rms or peak value instead of the instantaneous one in Question 1. Fix the phase $\omega t = \pi/3$ first.
• Thinking the generator frequency sets the resonant frequency in Question 4. It is set by $L$ and $C$ alone.
• Using $V^2/R$ or $V^2/Z$ for power in Question 6. The power is $I^2R$, or $VI\cos\phi$.
• Treating $24$ V as a peak value in Question 7. Meters and ratings give rms values.
• Missing that rise-then-fall means resonance in Question 8. It needs an inductor and a capacitor together.
• Choosing (c) in Question 10. A thinner line has more resistance and more loss.
• Reading $220$ V as the peak in Questions 12 and 13. It is the rms value; the peak is $220\sqrt2$ V and the mean is zero.

Practise next

• Exemplar Alternating Current: Very Short, Short and Long Answer Questions — where the same ideas are applied to equivalent circuits at high frequency, bandwidth, transformers and power loss in cables.
• NCERT textbook Chapter 7 — Alternating Current for the theory of rms values, phasors, resonance, power factor and the transformer.
• Exemplar Electromagnetic Induction — the previous unit, whose induced emf and self-inductance feed straight into $X_L$.
• Exemplar Electromagnetic Waves — the next unit, which carries the LC oscillator idea out into radiation.