 # 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?

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.

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?

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 $198.1 Pf$.