Question

In the given circuit, the voltage across load resistance (\( R_L \)) is:

• A. 8.75 V
• B. 9.00 V
• C. 8.50 V
• D. 14.00 V

Answer 1

The Correct Option is A

Given Data:

R_D1 = 1.5 kΩ,   R_L = 2.5 kΩ,   V = 15 V,   V_D1 = 0.3 V,   V_D2 = 0.7 V.

The voltage across the resistive portion of the circuit can be found by subtracting the voltage drops across the two diodes from the applied voltage:

V_resistive = V - (V_D1 + V_D2).

Substituting the values:

V_resistive = 15 V - (0.3 V + 0.7 V) = 14 V.

The total resistance in the resistive portion of the circuit is the sum of R_D1 and R_L:

R_total = R_D1 + R_L = 1.5 kΩ + 2.5 kΩ = 4.0 kΩ.

Using Ohm’s Law, the total current i in the circuit is given by:

i = \(\frac{V_{resistive}}{R_{total}}\).

Substituting the values:

i = \(\frac{14 \text{ V}}{4.0 \text{ k}\Omega}\) = \(\frac{14}{4000}\) A = 0.0035 A = 3.5 mA.

The voltage across the load resistance R_L, denoted as V_L, is given by Ohm’s Law:

V_L = i × R_L.

Substituting the values:

V_L = 3.5 mA × 2.5 kΩ = 0.0035 A × 2500 Ω = 8.75 V.

Final Result: The voltage across the load resistance R_L is:

V_L = 8.75 V