Question

An alternating voltage \(e = 150\sqrt{2}sin100t\) volt is applied to a capacitor of capacity \(2 μ\text{F}\). The root mean square value of current in the circuit is

• A. \(300\) mA
• B. \(150\) mA
• C. \(30\) mA
• D. \(15\) mA

Hint:

Find the capacitive reactance from omega and C, then divide the r.m.s. voltage by it.

Answer 1

The Correct Option is C

Step 1: Read the Voltage:
\(e=150\sqrt2\sin100t\) gives peak voltage \(150\sqrt2\) V and \(\omega=100\ \text{rad/s}\). The r.m.s. voltage is \(\dfrac{150\sqrt2}{\sqrt2}=150\) V.

Step 2: Capacitive Reactance:
\[ X_C=\frac1{\omega C}=\frac1{100\times2\times10^{-6}}=5000\ \Omega \]

Step 3: R.M.S. Current:
\[ I_{rms}=\frac{V_{rms}}{X_C}=\frac{150}{5000}=0.03\ \text{A}=30\ \text{mA} \]

Step 4: Check the Other Options:
300 mA and 150 mA would need \(X_C=500\ \Omega\) and \(1000\ \Omega\). 15 mA would need \(10{,}000\ \Omega\). Only 5000 ohm gives 30 mA.

Final Answer:
The r.m.s. current is 30 mA, option (C). \[ \boxed{\text{(C) } 30\ \text{mA}} \]