JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
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 and solution
Answer (a) 8.75 V
Solution
Physics sir · 3 steps
Take off the two diode drops first, then share what is left between the resistors in series.
- Both diodes conduct. Both arrows point along the current from the battery, so D₁ and D₂ are forward biased. A conducting Ge diode drops about 0.3 V and a Si diode about 0.7 V.
- Voltage left over. The resistors get 15 - 0.3 - 0.7 = 14 V. This is shared by 1.5 kΩ and R_L = 2.5 kΩ in series, a total of 4 kΩ.
I = (14 V)/(4 kΩ) = 3.5 mA
- So. V_L = I R_L = 3.5 mA × 2.5 kΩ = 8.75 V, which is option (a).
The idea to keep. In a series circuit with diodes, subtract each forward drop (Ge 0.3 V, Si 0.7 V) from the source first. Then use the voltage divider rule on the resistors.
Answer from the NTA final answer key. Solution by PrepWiser.