NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current
$$\textbf{Q. Show that the rms value of current is given by}\space I_0/\sqrt{2}\space\textbf{where}\space \textbf{I}_0\textbf{is the peak value of current.}$$
Ans. The rms value of current is given by
$$\text{I}^{2}\text{rms}=\frac{\int^{T}_{0} I^{2}dt}{\int^{T}_{0}dt}\\\text{But I=I}_0\space \text{sin}\omega t\\\text{I}^{2}_{\text{rms}}=\frac{I^{2}_0}{2T}\int^{T}_{0}\text{sin}^{2}\omega t\space dt\\=\frac{I^{2}_0}{2T}\int^{T}_{0}(1-\text{cos 2}\omega t)dt=\frac{I^{2}_0}{2T}\bigg(\int^{T}_{0}dt-\int^{T}_{0}\text{cos 2} \omega t\space dt \bigg)\\=\frac{I^{2}_{0}}{2T}×T=\frac{I^{2}_{0}}{2}\\\text{I}_{\text{rms}}=\frac{I_0}{\sqrt{2}}$$
Q. (i) For circuits used for transporting electric power, a low power factor implies large loss in transmission. Explain.
(ii) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
Ans. (i) As P = VI cos f where, cos f is the power factor. To supply a given power at a given voltage, if cos f is small we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission.
(ii) Suppose in a circuit, current I lags the voltage by an angle Φ. Then power factor cos Φ =
$$\text{cos}\phi=\frac{R}{Z}.$$
Q. A light bulb is rated at 100 W for a 220 V supply. Find :
- (i) the resistance of the bulb;
- (ii) the peak voltage of the source;
- (iii) the rms current through the bulb.
Ans. (i) We are given P = 100 W and V = 220 V. The resistance of the bulb is
$$\text{R} =\frac{V^{2}}{P}.\\=\frac{(220 V)^{2}}{100 W}=484Ω\\\text{(ii) The peak voltage of the source is}\\\text{V}_{m}=\sqrt{2}V=311\space V\\\text{(iii) Since, P=IV}\\ I=\frac{P}{V}=\frac{100 W}{220 V}=0.450A $$
Q. A 15.0 mF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (r.m.s. and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?
Ans. The capacitive reactance is
$$X_{c}=\frac{1}{2\pi fC}\\=\frac{1}{2\pi(50 Hz)(150×10^{-6}\text{F})}\\=212Ω\\\text{The r.m.s. current is,}\\i_{rms}=\frac{V}{X_C}=\frac{220V}{212Ω}=1.04A\\\text{The peak current is}\\\text{I}_0=\sqrt{2} i_{rms}=(1.41)(1.04 A)=1.47A\\\text{This current oscillates between + 1.47 A and – 1.47 A, and is ahead of the voltage by}\frac{\pi}{2}.\\\text{If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.}$$
Q. A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz a.c. supply.
- (i) What is the maximum current in the coil ?
- (ii) What is the time lag between the voltage maximum and the current maximum ?
Given, Inductance of inductor (L) = 0.50 H
Resistance of resistor (R) = 100 Ω
Potential of voltage supply (V) = 240 V
Frequency of the supply (n) = 50 Hz
$$\text{(i) Peak voltage is given by}\\(V_0)=\sqrt{2}V\\=\sqrt{2}×240=339.41V\\\text{Angular frequency (w)} = 2 \pi n = 2\pi × 50 = 100 \pi \text{rad/s}.\\\text{Maximum current in the circuit is given by}\\\text{I}_0=\frac{V_0}{\sqrt{R^{2}+\omega^{2}L^{2}}}\\\frac{339.41}{\sqrt{(100)^{2}+(100 \pi)^{2}(0.50)^{2}}}=1.82A\\\text{(ii)}\text{Equation for voltage is given by}\\\text{V = V}_0 \text{cos} \omega t\\\text{Equation for current is given by}\\\text{I = I}_0 \text{cos}(\omega t- \phi)\\\text{Here,}\phi=\text{Phase difference between voltage and current. At time t = 0.}\\\text{V = V}_0\text{(voltage is maximum)}\\\text{For}\space\omega t=\phi i.e,\text{at time (t) =}\frac{\phi}{\omega}.\\\text{I = I}_0 = \text{(current is maximum)}\\\text{Thus, the time lag between maximum voltage and maximum current is}\frac{\phi}{\omega}.\\\text{Now, phase angle}\phi\text{is given by the relation,}\\\text{tan}\phi=\frac{\omega L}{R}=\frac{2\pi×50×0.5}{100}\\=1.57\\\phi=57.5\degree=\frac{57.5\pi}{180}\text{rad}\\\omega t=\frac{57.5\pi}{180}\\\text{t}=\frac{57.5\pi}{180×2\pi×50}\\=3.19×10^{-3}s\\=3.2\text{ms}.\\\text{Therefore, the time lag between maximum voltage and maximum current is 3.2 ms.}$$
Q. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 mF is connected to a variable frequency 200 V a.c. 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?
Ans. At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
Given,
AC supply voltage to the LCR circuit, V = 200 V Impedance of the circuit is given by the relation,
$$\text{Z}=\sqrt{R^{2}+(X_L-X_C)^{2}}\\\text{At resonance, X}_L = X_C\\\therefore\text{Z}=R=20Ω\\\text{Current in the circuit can be calculated as :}\\\text{I}=\frac{V}{Z}=\frac{200}{20}=10A\\\text{Thus, the average power transferred to the circuit is one complete cycle :}\\\text{P=VI=200×10=2000\space W}$$
Q. A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 variable frequency supply.
- (i) What is the source frequency for which current amplitude is maximum ? Obtain this maximum value.
- (ii) What is the source frequency for which average power absorbed by the circuit is maximum ? Obtain the value of this maximum power.
- (iii) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency ? What is the current amplitude at these frequencies ?
- (iv) What is the Q-factor of the given circuit ?
- Ans. Given, Inductance (L) = 0.12 H
- Capacitance (C) = 480 nF = 480 × 10^{– 9 }F
- Resistance (R) = 23 Ω
- Supply voltage (V) = 230 V
- Peak voltages is given by
$$\text{(v)}_0=\sqrt{2}×230=325.22\text{V}\\\text{(i) The current flowing through the circuit is given by}\\\text{I}_0=\frac{V_0}{\sqrt{R^{2}+\bigg(\omega L-\frac{1}{\omega C}\bigg)^{2}}}\\\text{Here},\text{I}_0=\text{maximum at resonance.}\\\text{At resonance, we have}\\\omega_RL-\frac{1}{\omega _{R}C}=0\\\text{Here},\omega_R=\text{Resonance angular frequency}\\\therefore\omega_R=\frac{1}{\sqrt{LC}}\\=\frac{1}{\sqrt{0.12×480×10^{-9}}}\\=4166.67\space\text{rad/s}\\\therefore\text{Resonant frequency}(v_R)=\frac{\omega_R}{2\pi}\\=\frac{4166.67}{2×3.14}=663.48Hz\\\text{and Maximum current}\\\text{(I}_0)_{max}=\frac{V_0}{R}=\frac{325.22}{23}=14.14A\\\text{(ii) Maximum average power absorbed by the circuit is given as :}\\\text({P}_{av})_{max}=\frac{1}{2}(I_0)^{2}_{max}R\\=\frac{1}{2}×(14.14)^{2}×23=2299.3W\\\text{(iii)\text{The power transferred to the circuit is half the power at resonant frequency.}}\\\text{Frequencies for half transferred power}\\=\omega_R\pm\Delta\omega\\$$
$$= 2\pi (v_R \pm Δv)\\\text{Here},\Delta\omega=\frac{R}{2L}\\=\frac{23}{2×0.12}=95.83\space\text{rad/s}\\\text{Thus change in frequency,}\Delta v=\frac{1}{2\pi}\Delta\omega\\=\frac{95.83}{2\pi}=15.26Hz\\\therefore v_R+\Delta v=663.48+15 .26\\=678.74Hz\\\text{and}\space v_R-\Delta v=663.48+15.26\\=678.74Hz\\\text{and}\space v_R-\Delta v=663.48-15.26=648.22Hz\\\text{Hence},\text{at}\space648.22Hz\space\text{and}\space 678.74Hz\space\text{frequencies, the power transferred is half.}\\\text{At these frequencies, current amplitude can be given as :}\\\text{I}’=\frac{1}{\sqrt{2}}×(I_0)_{max}\\=\frac{14.14}{\sqrt{2}}=10A\\\text{(iv)\space\text{Q-factor of the given circuit can be obtained using the relation,}}\\\text{Q}=\frac{\omega_RL}{R}=\frac{4166.67×0.12}{23}\\=21.74\\\text{Hence, the Q-factor of the given circuit is 21.74.}$$
Q. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns.
What should be the number of turns in the secondary in order to get output power at 230 V?
- Ans. Given,
- Input voltage (V_{1}) = 2300V
- Number of turns in primary coil (n_{1}) = 4000
- Output voltage (V_{2}) = 230 V
- Number of turns in secondary coil = n_{2}
- Voltage is related to the number of turns as :
$$\frac{V_1}{V_2}=\frac{n_1}{n_2}\\\Rarr\frac{2300}{230}=\frac{4000}{n_2}\\\Rarr n_2=\frac{4000×230}{2300}=400\\\text{Hence, there are 400 turns in the second winding.}$$
Q. A light bulb and an open coil inductor are connected to an a.c. source through a key as shown in following fig.
The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor.
The glow of the light bulb : (i) increases, (ii) decreases, (iii) is unchanged, as the iron rod is inserted. Give your answer with reasons.
Ans. As soon as iron rod is inserted, the magnetic field inside the coil magnetizes the iron which increases the magnetic field inside it. Therefore, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. Hence, a larger fraction of the applied a.c. voltage appears across the inductor, leaving
less voltage across the bulb. Therefore, the glow of the light bulb decreases.
Q. A lamp is connected in series with a capacitor. Predict your observations for d.c. and a.c. connections. What happens in each case if the capacitance of the capacitor is reduced ?
Ans. When a d.c. source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With a.c. source, the capacitor offers capacitative reactance (1/wC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.
= 3.0 × 10^{– 5} × (π × 10^{– 2}) × cos 0°
= 3π × 10^{– 7} Wb
Therefore, estimated value of the induced emf is,
$$\epsilon=N\frac{\Delta\phi}{\Delta t}=\frac{500×(6\pi×10^{-7})}{0.25}\\=3.8×10^{-3}V\\\text{I}=\frac{\epsilon}{R}=\frac{3.8×10^{-3}}{2}=1.9×10^{-3}A\\\text{Note that the magnitude of} \epsilon \text{and I are the estimated values}.\\\text{Their instantaneous values are different and depend upon the speed of rotation at the particular instant.}$$
Q. Answer the following questions :
- (i) In any a.c. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit ? Is the same true for rms voltage ?
- (ii) A capacitor is used in the primary circuit of an induction coil.
- (iii) An applied voltage signal consists of a superposition of a d.c. voltage and an a.c. voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the d.c. signal will appear across C and the a.c. signal across L.
- (iv) A choke coil in series with a lamp is connected to a d.c. line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an a.c. line.
- (v) Why is choke coil needed in the use of fluorescent tubes with a.c. mains ? Why can we not use an ordinary resistor instead of the choke coil ?
- Ans. (i) Yes; the statement is not true for r.m.s. voltage.
- It is true that in any a.c. circuit, the applied voltage is equal to the average sum of the instantaneous voltages across the series elements of the circuit. However, this is not true for r.m.s. voltage because voltages across different elements may not be in phase.
- (ii) High induced voltage is used to charge the capacitor.
- A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.
- (iii) The d.c. signal will appear across capacitor C because for d.c. signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high almost infinite. Hence, a d.c. signal appears across C.
- For an a.c. signal of high frequency, the impedance of L is high and that of C is very low. Hence, an a.c. signal of high frequency appears across L.
- (iv) If an iron core is inserted in the choke coil (which is in series with a lamp connected to the a.c. line), then the lamp will glow dim. This is because the choke coil and the iron core increase the impedance of the circuit.
- (v) A choke coil is needed in the use of fluorescent tubes with a.c. mains because it reduces the voltage across the tube without wasting much power. An ordinary resistor cannot be used instead of a choke coil for this purpose because it wastes power in the form of heat.