An ideal diode is connected in series with a 1 kΩ load resistor and the input voltage is given as V(t) = sin²(t) + cos²(t) V. What is the average output voltage across the load resistor?

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RRB JE ECE 22 Apr 2025 Shift 2 CBT 2 Official Paper

Answer 1

Given:

Key Analysis:

Input Voltage Simplification:
Using the trigonometric identity:

sin2(t) + cos2(t) = 1

Therefore, V(t) = 1 V (constant DC voltage at all times)
Diode Behavior:
Since the input is always positive (1V):
- The diode remains continuously forward-biased
- Acts as a short circuit (0V drop)
Output Voltage:
The entire input appears across the resistor:

\( V_{out}(t) = V(t) = 1 \text{ V} \)

Average Voltage Calculation:

For a constant DC voltage:

\(V_{avg} = \frac{1}{T}\int_0^T 1\ dt = 1 \text{ V} \)

Final Answer:

4) +1 V

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