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Question: If a capacitor of \[8\mu F\] is connected to an \[220v,{\text{ }}100Hz\] ac source and the current p...

If a capacitor of 8μF8\mu F is connected to an 220v, 100Hz220v,{\text{ }}100Hz ac source and the current passing through it is65mA65mA, then the RMS voltage across it is

(A) 129.4V129.4V
(B) 12.94V12.94V
(C) 1.294V1.294V
(D) 15V15V

Explanation

Solution

Hint First, find the impedance due to the capacitance then find the value of the current in using the RMS value of the given source. The resulting value of the current will be in the RMS value only as we have used only the RMS value of the voltage.

Complete Step by step solution
Given the capacitance of a capacitor is C = 8μF=8×106  FC{\text{ }} = {\text{ }}8\mu F = 8 \times {10^{ - 6}}{\text{\;F}} since 1μ=1061\mu = {10^{ - 6}} units.
Also given the current passing through the capacitor is Irms=65mA=0.065A  {I_{rms}} = 65mA = 0.065A\;
The frequency of the source is given as v=100Hzv = 100Hz
We know that the capacitive reactance for a capacitor is represented by XC{X_C}
Therefore XC=1/2πvC{X_C} = 1/2\pi vC
By substituting the given values in the above equation we get
As XC=12×3.14×100×8×106=199{X_C} = \dfrac{1}{{2 \times 3.14 \times 100 \times 8 \times {{10}^{ - 6}}}} = 199
Since we know that the RMS voltage across the capacitor is
Vrms=IrmsXC=0.065×199=12.94  V{V_{rms}} = {I_{rms}}{X_C} = 0.065 \times 199 = 12.94\;V

Hence the correct option is B

Note The value of the impedance has to be found carefully after calculating the value of the frequency of the source. Then we need to find the value of the RMS voltage of the source voltage carefully. Finding these two values carefully can solve the problem easily.