Ncert Exercises & Solutions

If each diode in figure has a forward bias resistance of $25 Ω$ and infinite resistance in reverse bias, what will be the values of the currents $I_{1}, I_{2}, I_{3} and I_{4} ?$

Hint: Identify the series and parallel combinations.

Step 1: Find the equivalent resistance of the circuit.

Given,

Forward biased resistance = 25 Ω

Reverse biased resistance = \infty

As the diode in branch CD is reverse biased which is having infinite resistance,

so I_{3} = 0

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

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

AB is in parallel combination with EF.

So, resultant resistance \frac{1}{R'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} = \frac{1}{150} + \frac{1}{150} = \frac{2}{150}

\Rightarrow R' = 75 Ω

Total resistance R = R' + 25 = 75 + 25 = 100 Ω

Step 2 : Find the currents in the circuit .

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

I_{1} = I_{4} + I_{2} + I_{3} (Here I_{3} = 0)

So, I_{1} = I_{4} + I_{2}

Here, the resistance R1 and R2 is the same.

i . e ., I_{4} = I_{2}

∴ I_{1} = 2 I_{2}

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

and I_{4} = 0.025 A

Thus, I_{1} = 0.05 A, I_{2} = 0.025 A, I_{3} = 0.025 A, I_{3} = 0 and I_{4} = 0.025 A

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