Alternating Current

Physics

NCERT

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

Solution :

Given:$\\$ $R=100 \text{ohms}\\ V=220V$$\\$ Frequency(f) $=50 Hz$ $\\$ (a) We know $\\$ $I_{rms}=\dfrac{V_{rms}}{R}$$\\$ Substituting the values $\\$ $I_{rms}=\dfrac{220}{100}=2.20A$$\\$ (b)Power=$V.I$$\\$ Or Power=$220*2.2$$\\$ Or Power$=484 W$

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

Solution :

a) Peak voltage of the ac supply, $V _0 =300 V$$\\$ We know $V_{rms}=\dfrac{V_0}{\sqrt{2}}\\ =\dfrac{300}{\sqrt{2}}=212.1V$$\\$ b) The rms value of current is given as $I=10 A$$\\$ Using above identity for current peak current is given as:$\\$ $I_0=1.414*I_{rms}\\ \text{Or } I_0=1.414*10=14.14 A$

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

Solution :

Given:$\\$ $L = 44 mH = 44×10^{-3} HV = 220 V$$\\$ Frequency(f) $= 50 Hz$$\\$ Angular frequency,$\omega = 2 \pi f$$\\$ Inductive reactance,$X_L=\omega L = 2 \pi fL$$\\$ $I_{rms}$is given by =$\dfrac{V}{X_L}$$\\$ Determining inductive reactance$\\$ $X_L=2*3.14*50*44*10^{-3}\Omega\\ X_L=13.82 \text{ohms}$$\\$ Therefore $I_{rms}=220/13.82 \dfrac{220}{13.82}$$\\$ Or $ I_{rms}=15.92 A$

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

Solution :

Given:$\\$ C=$60$ microfarads$\\$ V=$110 $ volts$\\$ Frequency(f) $=60 Hz \ \ I_{rms}=\dfrac{V}{X_c}$$\\$ Now,$X_c=\dfrac{1}{2*3.14 * 60 * 60 * 10?^{(-6)}}\\ X_c=44.248 ohms $$\\$ Hence,$ I_{rms}=\dfrac{110}{44.248}=2.488 A$

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 :

In the inductive circuit,$\\$ Rms value of current, $I=15.92 A$$\\$ Rms value of voltage, $V=220 V$$\\$ Hence the net power absorbed can be obtained by the relation,$\\$ $P=VI \cos \phi$$\\$ Where,$\\$ $\phi=$ Phase difference between $V$ and $I$$\\$ For a pure inductive circuit, the phase difference alternating voltage and current is $90^0$$\\$ i.e., $\phi=90^0$$\\$ Hence,$P=0$ i.e., the net power is zero.$\\$ In the capacitive circuit,$\\$ rms value of current, $I=2.49 A$$\\$ rms value of voltage, $V=110 V$$\\$ Hence the net power absorbed can be obtained as:$\\$ $P=VI \cos \phi $$\\$ For a pure capacitive, the phase difference between alternating voltage and current is $90^0$ i.e.,$\phi=90^0$$\\$ Hence, $P=0,$ the net power is zero.

6   Obtain the resonant frequency $\omega _r$ of a series LCR circuit with $L = 2.0 H , C = 32 \mu F$ and $R = 10 \Omega $ . What is the Q-value of this circuit?

Solution :

Given:$\\$ $L = 2 H\\ C = 32 \mu F = 32×10 ^{-6} F\\ R = 10 \Omega $$\\$ Resonant frequency $ \omega_r=\dfrac{1}{\sqrt{LC}}$$\\$ Substitution yields $=\dfrac{1}{\sqrt{2*32*10^{-6}}}=\dfrac{1}{8 * 10^{-3}}$$\\$ $=125 \ rad /s $$\\$ Now Q-value $=\omega_ r L / R$$\\$ $Q=\dfrac{1}{R}\sqrt{L}{C}\\ =\dfrac{1}{10}\sqrt{\dfrac{2}{32*10^{-6}}}\\ =\dfrac{1}{10*4*10^{-3}}=25$$\\$ Hence, the Q-value is $=25$$\\$

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

Solution :

Given:$\\$ $C = 30 \mu F =30×10^{-6} F$$\\$ $L = 27 mH 27×10^{-3}$$\\$ Angular frequency of free oscillations $\omega _r=\dfrac{1}{\sqrt{LC}}$$\\$ $\dfrac{1}{\sqrt{27*10^{-3}*30*10^{-6}}}=\dfrac{1}{9*10^{-4}}=1.11*10^3 \ rad /s $$\\$ Hence, the angular frequency of free oscillations of the circuit is $=1.1*10^{3}s^{-1}$$\\$

8   Suppose the initial charge on the capacitor in Exercise $7.7$ is $6 mC.$. What is the total energy stored in the circuit initially? What is the total energy at later time?

Solution :

Capacitance of the capacitor, $C=30 \mu F = 30 * 10^{- 6} F$$\\$ Inductance of the inductor, $L=27mh=27×10 -3 H$$\\$ Charge on the capacitor,$Q=6Mc=6×10 -3 C$$\\$ $=E=\dfrac{1}{2}\dfrac{Q^2}{C}=\dfrac{1}{2}\dfrac{(6*10^{-3})^2}{30*10^{-6}}\\ =\dfrac{6}{10}=0.6 J$$\\$ Total energy at a later time will remain the same because energy is shared between the capacitor and the inductor.

9   A series LCR circuit with $R = 20 \Omega , L = 1.5 H $ and $ C =35 \mu F$ is connected to a variable- frequency $200 V$ 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 :

Given:$\\$ $R = 20 \Omega \\ L = 1.5 H\\ C = 35 \mu F = 30×10 -6 F\\ V = 200 volts $$\\$ Impedance of the circuit is given by the relation,$\\$ $Z=\sqrt{R^2+(X_L-X_C)^2}$$\\$ At resonance,$\\$ $X_L=X_C\\ Z=R=20 \Omega\\ So I=\dfrac{V}{Z}=\dfrac{200}{20}=10A $ Thus,$\\$ The average power transferred to the circuit in one complete cycle.$\\$ $P = I_ 2 R\\ Or P = 10 * 10 * 20 = 2000 W$$\\$

10   A radio can tune over the frequency range of a portion of MW broadcast band:$ (800 kHz$ to $1200kHz)$ .If its LC circuit has an effective inductance of $200 \mu H$ ,what must be the range of its variable capacitor? [Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.]

Solution :

The range of frequency (f) of aradio is $800KHZ$ to $1200KHZ$$\\$ Effective inductance of circuit $L=200 \mu H =200×10^{-6} H $$\\$ Capacitance of variable capacitor for $f_ 1$ is given as:$\\$ $C_1=\dfrac{1}{\omega_1^2 L}$ Where,$\omega_1=$ Angular frequency for capacitor for $f_1=2\pi f_1$$\\$ $2 \pi * 800 * 10^3 \ rad /s\\ \therefore C_1=\dfrac{1}{(2 \pi * 800 * 10^3)^2* 200 * 10^{-6}}\\ =1.9809*10^{-10}F=198 pf\\ C_2=\dfrac{1}{\omega_2^2 L}\\ \therefore C_2=\dfrac{1}{(2\pi * 1200 * 10^3)^2* 200 * 10^{-6}}\\ =0.8804*10^{-10}F=88pF$$\\$ Hence, the range of the variable capacitor is from $88.04 pF$ to $198.1 Pf$.