Question:
\(\dfrac{50}{\pi}\ \mu\text{F}\) capacitor is connected to a \(250\ \text{V},\ 50\ \text{Hz}\) AC supply. Then the rms current of the circuit is:
\(\dfrac{50}{\pi}\ \mu\text{F}\) capacitor is connected to a \(250\ \text{V},\ 50\ \text{Hz}\) AC supply. Then the rms current of the circuit is:
Show Hint
For a pure capacitive AC circuit,
\[
I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C}
\]
where
\[
X_C=\frac{1}{\omega C}
\]
Higher capacitance results in lower reactance and hence larger current.
Updated On: Jun 26, 2026
- \(1.25\ \text{A}\)
- \(4.9\ \text{A}\)
- \(5\ \text{A}\)
- \(6\ \text{A}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the formula for capacitive reactance.
The capacitive reactance is \[ X_C=\frac{1}{\omega C} \] where \[ \omega=2\pi f \] Given, \[ f=50\ \text{Hz} \] Thus, \[ \omega=2\pi(50) \] \[ \omega=100\pi\ \text{rad s}^{-1} \]
Step 2: Convert the capacitance into SI units.
Given capacitance: \[ C=\frac{50}{\pi}\ \mu\text{F} \] Since \[ 1\ \mu\text{F}=10^{-6}\ \text{F} \] Therefore, \[ C=\frac{50\times10^{-6}}{\pi}\ \text{F} \]
Step 3: Calculate the capacitive reactance.
Substituting the values, \[ X_C= \frac{1}{ (100\pi) \left( \frac{50\times10^{-6}}{\pi} \right) } \] Cancelling \(\pi\), \[ X_C= \frac{1}{100\times50\times10^{-6}} \] \[ X_C= \frac{1}{5\times10^{-3}} \] \[ X_C=200\ \Omega \]
Step 4: Calculate the rms current.
Using Ohm’s law for AC circuits, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C} \] Given, \[ V_{\text{rms}}=250\ \text{V} \] Therefore, \[ I_{\text{rms}}=\frac{250}{200} \] \[ I_{\text{rms}}=1.25\ \text{A} \]
Step 5: Final conclusion.
Therefore, the rms current in the circuit is \[ \boxed{1.25\ \text{A}} \]
The capacitive reactance is \[ X_C=\frac{1}{\omega C} \] where \[ \omega=2\pi f \] Given, \[ f=50\ \text{Hz} \] Thus, \[ \omega=2\pi(50) \] \[ \omega=100\pi\ \text{rad s}^{-1} \]
Step 2: Convert the capacitance into SI units.
Given capacitance: \[ C=\frac{50}{\pi}\ \mu\text{F} \] Since \[ 1\ \mu\text{F}=10^{-6}\ \text{F} \] Therefore, \[ C=\frac{50\times10^{-6}}{\pi}\ \text{F} \]
Step 3: Calculate the capacitive reactance.
Substituting the values, \[ X_C= \frac{1}{ (100\pi) \left( \frac{50\times10^{-6}}{\pi} \right) } \] Cancelling \(\pi\), \[ X_C= \frac{1}{100\times50\times10^{-6}} \] \[ X_C= \frac{1}{5\times10^{-3}} \] \[ X_C=200\ \Omega \]
Step 4: Calculate the rms current.
Using Ohm’s law for AC circuits, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{X_C} \] Given, \[ V_{\text{rms}}=250\ \text{V} \] Therefore, \[ I_{\text{rms}}=\frac{250}{200} \] \[ I_{\text{rms}}=1.25\ \text{A} \]
Step 5: Final conclusion.
Therefore, the rms current in the circuit is \[ \boxed{1.25\ \text{A}} \]
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