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.