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

• A. \(0.05\,\text{A} \)
• B. \(0.07\,\text{A} \)
• C. \(0.02\,\text{A} \)
• D. \(0.03\,\text{A} \)
• E. \(0.04\,\text{A} \)

Hint:

Use peak voltage \( V_{rms}\sqrt{2} \) in AC calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Input Voltage:
When an AC voltage is specified without qualification (like "peak" or "peak-to-peak"), it is always the Root Mean Square (RMS) value.
So, the given RMS value of the input voltage is \(V_{rms} = 250\) V.
Step 2: Calculating the Peak Input Voltage:
The peak voltage (\(V_{peak}\) or \(V_0\)) is related to the RMS voltage by the formula:
\[ V_{peak} = V_{rms} \times \sqrt{2} \] \[ V_{peak} = 250 \times \sqrt{2} \approx 250 \times 1.414 = 353.5 \, \text{V} \] Step 3: Analyzing the Circuit Resistance:
In a full wave rectifier, during any given half-cycle, the current flows through one of the diodes and then through the load resistor. Therefore, the total resistance in the circuit at any instant is the sum of the internal resistance of the conducting diode and the load resistance.
Internal resistance of diode, \(R_{diode} = 10 \, \Omega\).
Load resistance, \(R_{load} = 5 \, \text{k}\Omega = 5000 \, \Omega\).
Total resistance, \(R_{total} = R_{diode} + R_{load} = 10 + 5000 = 5010 \, \Omega\).
Step 4: Calculating the Peak Output Current:
The peak value of the output current (\(I_{peak}\)) will occur when the input voltage is at its peak. Using Ohm's law:
\[ I_{peak} = \frac{V_{peak}}{R_{total}} \] \[ I_{peak} = \frac{353.5 \, \text{V}}{5010 \, \Omega} \approx 0.07056 \, \text{A} \] This value is approximately 0.07 A.
Step 5: Final Answer:
The peak value of the output current is approximately 0.07 A.