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: (a). Both diodes are forward biased: drop across Ge = 0.3 V, across Si = 0.7 V. I = (15 - 0.3 - 0.7)/(1.5 + 2.5) kΩ = (14)/(4000) = 3.5 mA VL = I RL = 3.5× 10⁻³× 2500 = 8.75 V So the answer is (a) 8.75 V.

In the given circuit, the voltage across load resistance (R L) is: | JEE Main 2024 (30 Jan, Shift 2)

Q49JEE Main 2024 · 30 JAN · SHIFT 2Medium

P-N Junction Diode

In the given circuit, the voltage across load resistance (R ${}_L$ ) is:

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Answer: (a)

Both diodes are forward biased: drop across Ge $= 0.3$ V, across Si $= 0.7$ V.
$I = \dfrac{15 - 0.3 - 0.7}{1.5 + 2.5}\text{ k}\Omega = \dfrac{14}{4000} = 3.5$ mA
$V_L = I R_L = 3.5\times 10^{-3}\times 2500 = 8.75$ V
So the answer is (a) $8.75$ V.

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