Q&A - Ask Doubts and Get Answers
Q

A 100 μF capacitor in series with a 40 omega resistance is connected to a 110 V ,60 Hz supply.

Q7.15    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)
Views

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.

Exams
Articles
Questions