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

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

Answers (1)

Explanation:-

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

As shown in figure,e the diode in branch CD is in reverse biases which have infinite resistance

So, the 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 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 are the 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

Posted by

infoexpert24

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