Semiconductors

JEE Main level, single correct, medium

46 of 209

Question 46 of 209

In the given circuit, the voltage across load resistance (RL)({R}_{L}) is:

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Answer: (A) 8.75V8.75V


Both diodes are in forward bias. In forward bias, potential drop across GeGe and SiSi will be 0.3V&0.7V respectively.
Therefore, current in the circuit i=15−0.3−0.74×1000=3.5mAi=\dfrac{15-0.3-0.7}{4\times 1000}=3.5mA
Required potential drop across RL{R}_{L},
VL=iRL=(3.5×10−3)×(2.5×1000)V{V}_{L}=i{R}_{L}=\left(3.5\times {10}^{-3}\right)\times \left(2.5\times 1000\right)V
=8.75V=8.75V

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