Semiconductors

JEE Main level, single correct, medium

50 of 209

Question 50 of 209

If each diode has a forward bias resistance of 25Ω25\Omega in the below circuit,

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Answer: (C) I1I2=2\dfrac{{I}_{1}}{{I}_{2}}=2

As the diode D2{D}_{2} is connected in reverse order, there is no current through it.
The effective resistances of the circuit can, then, be calculated as follows:
Reff=25+(25+125)2(25+125)+(25+125)Ω=100Ω\begin{matrix}{R}_{eff} & = & 25+\dfrac{{\left(25+125\right)}^{2}}{\left(25+125\right)+\left(25+125\right)}\Omega \\ & = & 100\Omega \end{matrix}
Each current through the diodes can be calculated as follows:
I1=(5V100Ω)=0.05A\begin{matrix}{I}_{1} & = & \left(\dfrac{5V}{100\Omega }\right) \\ & = & 0.05A\end{matrix}
As the diode D2{D}_{2} is non-conducting, the current I1{I}_{1} will be divided equally among the other two diodes D1{D}_{1} and D3{D}_{3}.
Hence, the currents in the diodes can be written as
I2=0.05A2=0.025AI4=0.025A\begin{matrix}{I}_{2} & = & \dfrac{0.05A}{2} \\ & = & 0.025A \\ {I}_{4} & = & 0.025A\end{matrix}
Thus,
I1I2=0.05A0.025A=2\begin{matrix}\dfrac{{I}_{1}}{{I}_{2}} & = & \dfrac{0.05A}{0.025A} \\ & = & 2\end{matrix}

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