Question

A $40\, \mu\,F$ capacitor is connected to a $200\,V$,$50\,Hz$ ac supply. The rms value of the current in the circuit is, nearly :

• A. 1. 7 A
• B. 2. 05 A
• C. 2. 5 A
• D. 25. 1 A

Answer 1

The Correct Option is C

To find the rms value of the current in the circuit, we first need to understand the concept of capacitive reactance in an AC circuit.

Capacitive Reactance (\(X_c\)):

When a capacitor is connected to an AC supply, it exhibits capacitive reactance, which is given by the formula:

\(X_c = \dfrac{1}{2\pi f C}\)

where:

• \(f\) is the frequency of the AC supply,
• \(C\) is the capacitance.

Given Values:

• \(C = 40\, \mu F = 40 \times 10^{-6}\, F\)
• \(V = 200\, V\)
• \(f = 50\, Hz\)

Step 1: Calculate Capacitive Reactance (\(X_c\))

Substitute the given values into the formula for capacitive reactance:

\(X_c = \dfrac{1}{2\pi \times 50 \times 40 \times 10^{-6}}\)

After calculation, we get:

\(X_c \approx 79.58\, \Omega\)

Step 2: Calculate the RMS Current (\(I_{rms}\))

The rms current in the circuit can be described using Ohm's Law for AC circuits:

\(I_{rms} = \dfrac{V_{rms}}{X_c}\)

Substitute the given values:

\(I_{rms} = \dfrac{200}{79.58}\)

After calculation, we find:

\(I_{rms} \approx 2.51\, A\)

Thus, the closest option is 2. 5 A.