Question:
An alternating voltage of \( 250\,\text{V},\,50\,\text{Hz} \) is applied to a full wave rectifier. If the internal resistance of each diode is \( 10\,\Omega \) and the load resistance is \( 5\,\text{k}\Omega \), the peak value of output current is
An alternating voltage of \( 250\,\text{V},\,50\,\text{Hz} \) is applied to a full wave rectifier. If the internal resistance of each diode is \( 10\,\Omega \) and the load resistance is \( 5\,\text{k}\Omega \), the peak value of output current is
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Use peak voltage \( V_{rms}\sqrt{2} \) in AC calculations.
Updated On: Apr 21, 2026
- \(0.05\,\text{A} \)
- \(0.07\,\text{A} \)
- \(0.02\,\text{A} \)
- \(0.03\,\text{A} \)
- \(0.04\,\text{A} \)
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The Correct Option is B
Solution and Explanation
Concept:
\[
I_{peak} = \frac{V_{peak}}{R}
\]
Step 1: Peak voltage.
\[ V_{peak} = 250\sqrt{2} \]
Step 2: Total resistance.
\[ R = 5000 + 2\times10 = 5020\,\Omega \]
Step 3: Current.
\[ I \approx \frac{250\sqrt{2}}{5020} \approx 0.07\,\text{A} \]
Step 1: Peak voltage.
\[ V_{peak} = 250\sqrt{2} \]
Step 2: Total resistance.
\[ R = 5000 + 2\times10 = 5020\,\Omega \]
Step 3: Current.
\[ I \approx \frac{250\sqrt{2}}{5020} \approx 0.07\,\text{A} \]
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