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If each diode in Fig. 14.13 has a forward bias resistance of 25\Omega and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4?

Answers (1)

According to the problem forward biases resistance = 25 \Omega and reverse biases resistance=\infty

As show in figure the diode in brsnch CD is in reverse biases which having infinite resistance

So current in that branch is zero i.e I_{3}=\theta

Resistance in branch AB=25+125=150\Omega\ say \ R_{1}

Resistance in branch EF=25+125=150\Omega say R_{2}

AB is parallel to EF

So effective resistance

\frac{1}{R'}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}\\\Rightarrow R'=75\Omega

Total resistance R of the circuit=R'=25=75+25=100\Omega



According to the Kirchhoff's, current law (KCL)

I_{1}=I_{4}+I_{2}+I_{3}\: \: \: \: \: \: \: \left ( I_{3}=0 \right )\\ SoI_{1}=I_{4}+I_{2}

Here the resistances R1 and R2 is same


  I_{4}=I_{2}\\ \therefore I_{1}=2I_{2}\\ \Rightarrow I_{2}=\frac{I_{1}}{2}=\frac{0.05}{2}=0.025A

And I4 =0.025A

Therefore we get I1 =0.05A,I2 =0.025A,I3 =0 and I4 =0.025A

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