NCERT Class 12 Physics — Alternating Current, Chapter 7 Exercises. All 8 questions solved.

The exercises at the end of Chapter 7 build up the series $LCR$ circuit one element at a time. Questions 7.1 and 7.2 are about rms values in a purely resistive circuit, 7.3 to 7.5 put a single inductor or capacitor across the supply, 7.6 is the $LC$ circuit left to oscillate on its own, and 7.7 and 7.8 are the full series $LCR$ circuit at resonance. The arithmetic runs on three sets of results:

• rms values — $I = I_m/\sqrt2$ and $V = V_m/\sqrt2$. A supply described as “220 V” means $220\ \text{V}$ rms, and ac meters read rms values.
• Reactance — an inductor opposes current with $X_L = \omega L$ and a capacitor with $X_C = 1/\omega C$, where $\omega = 2\pi\nu$ is the angular frequency.
• Series $LCR$ circuit — impedance $Z = \sqrt{R^2 + (X_C – X_L)^2}$, resonance at $\omega_0 = 1/\sqrt{LC}$, and average power $P = VI\cos\phi$, where $\phi$ is the phase difference between voltage and current.

Key insight. Keep track of the phase between current and voltage, because it decides everything else here. In a resistor they are in step, so power is drawn every instant (7.1). In a pure inductor or capacitor they are a quarter-cycle apart, so the power drawn in one half-cycle is handed back in the next and the average is zero (7.5). At resonance the inductor’s and capacitor’s voltages are exactly opposite and cancel, so the circuit behaves as if only the resistor were there (7.7, 7.8).

Question 7.1

A $100\ \Omega$ resistor is connected to a $220\ \text{V}$, $50\ \text{Hz}$ ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle?

Question 7.1 (a)

Solution. The $220\ \text{V}$ quoted for a supply is its rms value. In a resistor, current and voltage are in phase and Ohm’s law holds for rms values just as it does for dc:

$$I = \frac{V}{R} = \frac{220}{100} = 2.20\ \text{A}$$

Question 7.1 (b)

Solution. Averaged over a full cycle, the power in a resistor is the product of the rms values. That is what rms values are defined to do: they give the same average heating as a direct current of that size. Here $\cos\phi = 1$ because current and voltage are in phase:

$$P = VI = (220)(2.20) = 484\ \text{W}$$

Question 7.2

(a) The peak voltage of an ac supply is $300\ \text{V}$. What is the rms voltage? (b) The rms value of current in an ac circuit is $10\ \text{A}$. What is the peak current?

Question 7.2 (a)

Solution. For a sinusoidal voltage the rms value is the peak value divided by $\sqrt2$:

$$V = \frac{V_m}{\sqrt2} = \frac{300}{\sqrt2} = 212.1\ \text{V}$$

Question 7.2 (b)

Solution. The same relation run backwards: the peak is $\sqrt2$ times the rms value.

$$I_m = \sqrt2\,I = 10\sqrt2 = 14.1\ \text{A}$$

Question 7.3

A $44\ \text{mH}$ inductor is connected to a $220\ \text{V}$, $50\ \text{Hz}$ ac supply. Determine the rms value of the current in the circuit.

Solution. An inductor opposes a changing current by its self-induced emf, and the opposition it offers is its inductive reactance, which grows with frequency:

$$X_L = \omega L = 2\pi\nu L = 2\pi(50)(44\times10^{-3}) = 13.8\ \Omega$$

The reactance plays the part of resistance in Ohm’s law for rms values:

$$I = \frac{V}{X_L} = \frac{220}{13.8} = 15.9\ \text{A}$$

Question 7.4

A $60\ \mu\text{F}$ capacitor is connected to a $110\ \text{V}$, $60\ \text{Hz}$ ac supply. Determine the rms value of the current in the circuit.

Solution. A capacitor lets alternating current flow because it is charged and discharged every half-cycle. Its opposition is the capacitive reactance, which falls as the frequency rises:

$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi(60)(60\times10^{-6})} = 44.2\ \Omega$$

$$I = \frac{V}{X_C} = \frac{110}{44.2} = 2.49\ \text{A}$$

Question 7.5

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Solution. The average power is $P = VI\cos\phi$. In a pure inductor the current lags the voltage by $\pi/2$, and in a pure capacitor it leads by $\pi/2$; either way $\cos\phi = \cos(\pi/2) = 0$. Physically, during one quarter-cycle the inductor stores energy in its magnetic field (or the capacitor in its electric field), and during the next quarter-cycle it returns all of it to the source. Energy moves back and forth, but none is used up. So although currents of $15.9\ \text{A}$ and $2.49\ \text{A}$ flow, neither circuit absorbs any net power.

Question 7.6

A charged $30\ \mu\text{F}$ capacitor is connected to a $27\ \text{mH}$ inductor. What is the angular frequency of free oscillations of the circuit?

Solution. The capacitor discharges through the inductor, the inductor’s magnetic field then recharges the capacitor the other way, and the energy swings between the two at the natural angular frequency of the $LC$ circuit:

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(27\times10^{-3})(30\times10^{-6})}} = \frac{1}{\sqrt{8.1\times10^{-7}}} = \frac{1}{9\times10^{-4}} = 1.1\times10^3\ \text{rad s}^{-1}$$

This corresponds to a frequency $\nu_0 = \omega_0/2\pi \approx 177\ \text{Hz}$.

Question 7.7

A series $LCR$ circuit with $R = 20\ \Omega$, $L = 1.5\ \text{H}$ and $C = 35\ \mu\text{F}$ is connected to a variable-frequency $200\ \text{V}$ ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Solution. When the supply frequency equals the natural frequency, the circuit is at resonance: $X_L = X_C$, the two reactances cancel, and the impedance falls to its smallest possible value, $Z = R$. The values of $L$ and $C$ are therefore not needed. The current and voltage are in phase, so $\cos\phi = 1$. Taking the $200\ \text{V}$ as rms:

$$I = \frac{V}{R} = \frac{200}{20} = 10\ \text{A}, \qquad P = I^2R = (10)^2(20) = 2000\ \text{W}$$

Question 7.8

Figure 7.17 shows a series $LCR$ circuit connected to a variable frequency $230\ \text{V}$ source. $L = 5.0\ \text{H}$, $C = 80\ \mu\text{F}$, $R = 40\ \Omega$. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the $LC$ combination is zero at the resonating frequency.

Question 7.8 (a)

Solution. Resonance happens when the inductive and capacitive reactances are equal, $\omega L = 1/\omega C$, which gives

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(5.0)(80\times10^{-6})}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{0.02} = 50\ \text{rad s}^{-1}$$

The corresponding frequency in hertz is $\nu_0 = \omega_0/2\pi = 50/2\pi = 7.96\ \text{Hz}$.

Question 7.8 (b)

Solution. At resonance $X_L = X_C$, so the reactive term in the impedance vanishes:

$$Z = \sqrt{R^2 + (X_C – X_L)^2} = R = 40\ \Omega$$

The source’s $230\ \text{V}$ is an rms value, so the rms current is $230/40 = 5.75\ \text{A}$. The question asks for the amplitude, which is $\sqrt2$ times the rms value:

$$I_0 = \sqrt2 \times 5.75 = 8.13\ \text{A}$$

Question 7.8 (c)

Solution. Each element’s rms voltage is the rms current, $5.75\ \text{A}$, times that element’s resistance or reactance. At $\omega_0 = 50\ \text{rad s}^{-1}$ the reactances are $X_L = \omega_0L = 250\ \Omega$ and $X_C = 1/\omega_0C = 1/(50\times80\times10^{-6}) = 250\ \Omega$ — equal, as resonance requires. So

$$V_R = (5.75)(40) = 230\ \text{V}, \qquad V_L = (5.75)(250) = 1437.5\ \text{V}, \qquad V_C = (5.75)(250) = 1437.5\ \text{V}$$

The voltages across $L$ and $C$ are each more than six times the supply voltage, yet they do not add to it. The voltage across the inductor leads the current by $\pi/2$ and the voltage across the capacitor lags it by $\pi/2$, so the two are exactly opposite at every instant. Across the $LC$ combination, then,

$$V_{LC} = I\left(\omega_0L – \frac{1}{\omega_0C}\right) = 5.75\,(250 – 250) = 0$$

and the whole supply voltage appears across the resistor, which is why $V_R = 230\ \text{V}$.

Common mistakes

• Question 7.1(b): using peak values, or halving the power. The $220\ \text{V}$ is already rms, so $P = VI = 484\ \text{W}$ directly. Students who treat it as a peak value divide by 2 and get $242\ \text{W}$.
• Question 7.3: forgetting the $2\pi$ in the reactance. $X_L = \omega L$ uses the angular frequency, $\omega = 2\pi\nu = 314\ \text{rad s}^{-1}$ here. Writing $X_L = \nu L = 2.2\ \Omega$ gives a current $2\pi$ times too large, $100\ \text{A}$.
• Question 7.4: inverting the capacitive reactance. $X_C = 1/\omega C$ falls as the frequency or the capacitance rises. Writing $X_C = \omega C$ gives $0.023\ \Omega$ and an absurd current of nearly $5000\ \text{A}$.
• Question 7.5: answering $VI$ because a current flows. A current certainly flows, but it is a quarter-cycle out of step with the voltage, so $\cos\phi = 0$. The energy stored in one quarter-cycle is returned in the next.
• Question 7.6: quoting $\omega_0$ in hertz. $1/\sqrt{LC}$ is an angular frequency, in $\text{rad s}^{-1}$. The frequency in hertz is smaller by a factor of $2\pi$, about $177\ \text{Hz}$.
• Question 7.7: using $L$ and $C$ to find the impedance. At resonance the reactances cancel and $Z = R$. Only the resistance and the supply voltage are needed.
• Question 7.8(b): giving the rms current when the amplitude is asked for. $230/40 = 5.75\ \text{A}$ is the rms current. The amplitude is $\sqrt2$ times that, $8.1\ \text{A}$.
• Question 7.8(c): expecting $V_R + V_L + V_C = 230\ \text{V}$. Voltages in an ac circuit add as phasors, not as numbers. $V_L$ and $V_C$ are opposite in phase and cancel, which is how each can be $1437.5\ \text{V}$ across a $230\ \text{V}$ supply.

Practise next

• Chapter 8, Electromagnetic Waves — the next chapter starts from the capacitor in an ac circuit and asks what current flows between its plates.
• Chapter 6, Electromagnetic Induction — the self-induced emf behind every inductive reactance in this chapter.