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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?

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$\Omega$

current

$I_{1}=\frac{V}{R}=\frac{5}{100}=0.05A$

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.e

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

And I4 =0.025A

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