Q7.15 (a)   100\mu F capacitor in series with a 40\Omega resistance is connected to a 110\: V, 60\: Hzsupply.
 What is the maximum current in the circuit?

Answers (1)

Given,

The capacitance of the capacitor C=100\mu F

The resistance of the circuit R=40\Omega

Voltage supply V = 100V

Frequency of voltage supply f=60Hz

The maximum current in the circuit

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *60*100*10^{-6}} \right )^2}}=3.24A

Hence maximum current in the circuit is 3.24A.

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