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

Answer: (A)

8.75 V

Answer 2

Given Data:

RD1 = 1.5 kΩ, RL = 2.5 kΩ, V = 15 V, VD1 = 0.3 V, VD2 = 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:

Vresistive = V - (VD1 + VD2).

Substituting the values:

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

The total resistance in the resistive portion of the circuit is the sum of RD1 and RL:

Rtotal = RD1 + RL = 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 RL, denoted as VL, is given by Ohm’s Law:

VL = i × RL.

Substituting the values:

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

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

VL = 8.75 V