An a.c. voltage of peak value 20 V is connected in series with a silicon diode andload resistance of 500 Ω. If the forward resistance of diode is 10 Ω, find :

1. Break Down the Problem

To find the output voltage $V_{out}$ across the load resistance when an AC voltage is applied with a silicon diode in series, we'll consider:

1. The peak voltage of the AC source.
2. The voltage drop across the silicon diode.
3. The total resistance in the circuit (load resistance plus diode's forward resistance).
4. The output voltage across the load resistance.

2. Relevant Concepts

1. The peak value of the input voltage ($V_{peak}$): $20 \, V$.
2. The forward voltage drop of a silicon diode ($V_{diode}$): typically about $0.7 \, V$.
3. Ohm’s Law: $V = I \times R$.

3. Analysis and Detail

1. Calculate the voltage across the diode: $V_{diode} = 0.7 \, V$

2. Calculate the remaining voltage across the load resistance: Since the diode and load resistance are in series, the output voltage ($V_{out}$) can be calculated as: $V_{remaining} = V_{peak} - V_{diode} = 20 \, V - 0.7 \, V = 19.3 \, V$

3. Calculate the total resistance in the circuit: The total resistance $R_{total}$ is the sum of the load resistance ($R_L$) and the diode forward resistance ($R_{diode}$): $R_{total} = R_L + R_{diode} = 500 \, \Omega + 10 \, \Omega = 510 \, \Omega$

4. Calculate the current through the circuit using Ohm's law: $I = \frac{V_{remaining}}{R_{total}} = \frac{19.3 \, V}{510 \, \Omega} \approx 0.03784 \, A$

5. Calculate the voltage across the load resistance ($V_L$): Use Ohm's Law again: $V_L = I \times R_L = 0.03784 \, A \times 500 \, \Omega \approx 18.92 \, V$

4. Verify and Summarize

1. The calculated current $I$ is used to find the voltage across the load.
2. All steps are verified, and the values used are appropriate.

Final Answer

The voltage across the load resistance is approximately $18.92 \, V$.