In an a.c. circuit an alternating voltage $e=200 \sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity 1$\mu$F. The r.m.s. value of the current in the circuit is:
A. 10 mA
B. 100 mA
C. 200 mA
D. 20 mA

Question

In an a.c. circuit an alternating voltage $e=200 \sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity 1$\mu$F. The r.m.s. value of the current in the circuit is:
A. 10 mA
B. 100 mA
C. 200 mA
D. 20 mA

Vedantu Content Team · Accepted Answer

Hint: The emf of the supply voltage is given in the question. Like a resistor, a capacitor also creates some voltage drop across. It has some 'impedance', which when multiplied by current gives the voltage drop across it.Formula Used:Capacitor's reactance (or resistance) is obtained as:$X_C = \dfrac{1}{\omega C}$Complete step by step answer:Given, the supply:$e=200 \sqrt{2} \sin 100t$ volts And the value of C = 1$\mu$ FWe know that an alternating current source will have a supply voltage given by $V_0 \sin \omega t$ type of expression. We only compare this with our given supply voltage. The direct comparison gives us:$V_0 = 200 \sqrt{2}$ voltsand$\omega = 100$ HzThe reactance of the capacitor can be found by plugging in the values as:$X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 	imes 10^{-6}} = 10000$ ohmsWe know, by ohm's law that the current I in the circuit with resistance R and supply voltage V is just V/R. Similarly, we can write:$I_0 = \dfrac{V_0}{X_C}$Thus, $I_0 = \dfrac{200 \sqrt{2}}{10000}$ABut, we need to find the rms value of the current, so we use the formula:$I_{rms} = \dfrac{I_0}{\sqrt{2}}$Therefore, we can get:$I_{rms} = \dfrac{2}{100}$AOr, we can say:$I_{rms} = 20$mAThus, the correct option (D). 20 mA. This is the value of rms ac current in the circuit consisting of 1$\mu$F capacitor.Note:If more components are present in the circuit, the situation becomes different. If two capacitors were present in series we had to use equivalent capacitance in the reactance formula. If a resistor would have been present we would have done some vector algebra as the current in a capacitor lags behind from a resistor.

In an a.c. circuit an alternating voltage $e=200 \sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity 1$\mu$F. The r.m.s. value of the current in the circuit is:
A. 10 mA
B. 100 mA
C. 200 mA
D. 20 mA

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In an a.c. circuit an alternating voltage $e=200 \sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity 1$\mu$F. The r.m.s. value of the current in the circuit is:A. 10 mAB. 100 mAC. 200 mAD. 20 mA

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In an a.c. circuit an alternating voltage $e=200 \sqrt{2} \sin 100t$ volts is connected to a capacitor of capacity 1$\mu$F. The r.m.s. value of the current in the circuit is:
A. 10 mA
B. 100 mA
C. 200 mA
D. 20 mA