An a.c. voltage of peak value 20 V is connected in series with a silicon diode and a load resistance of 500 G. The forward resistance of the diode is 10 G and the barrier voltage is 0.7 V. Find the peak current through diode and the peak voltage across the load.

Options

A. 37.8 mA, 18.9 V
B. 5 mA, 30 V
C. 30 mA, 5 V
D. 2 mA, 2 V

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Circuit
We have an AC voltage source with a peak value of \( V_p = 20 \, V \), a silicon diode with a forward resistance \( R_d = 10 \, \Omega \), a barrier voltage \( V_b = 0.7 \, V \), and a load resistance \( R_L = 500 \, \Omega \).

### Step 2: Calculate the Total Resistance
The total resistance in the circuit when the diode is forward-biased is the sum of the load resistance and the forward resistance of the diode:
\[ R_{total} = R_L + R_d = 500 \, \Omega + 10 \, \Omega = 510 \, \Omega \]

### Step 3: Calculate the Effective Voltage Across the Circuit
The effective voltage across the diode and the load resistance is the peak voltage minus the barrier voltage:
\[ V_{effective} = V_p - V_b = 20 \, V - 0.7 \, V = 19.3 \, V \]

### Step 4: Calculate the Peak Current Through the Diode
Using Ohm's law, the peak current \( I_p \) through the diode can be calculated as:
\[ I_p = \frac{V_{effective}}{R_{total}} = \frac{19.3 \, V}{510 \, \Omega} \approx 0.0378 \, A = 37.8 \, mA \]

### Step 5: Calculate the Peak Voltage Across the Load
The peak voltage across the load resistance \( V_L \) can be calculated using Ohm's law:
\[ V_L = I_p \times R_L = 0.0378 \, A \times 500 \, \Omega \approx 18.9 \, V \]

### Final Results
- The peak current through the diode is approximately \( 37.8 \, mA \).
- The peak voltage across the load is approximately \( 18.9 \, V \).