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  • Currents flowing in each of the circuits A and B respectively are <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/20/p15.png" style="height:227px; width:564px" /

Currents flowing in each of the circuits A and B respectively are

Option: 1

1 A, 2 A


Option: 2

2 A, 1 A


Option: 3

4 A, 2 A


Option: 4

2 A, 4 A


Answers (1)

best_answer

In circuit A, both (p-n)junction diode act as forward biased. Hence, current flows in circuit A,
Total resistance Ris given by
\mathrm{\frac{1}{R}=\frac{1}{4}+\frac{1}{4} \Rightarrow \frac{1}{R}=\frac{2}{4} \Rightarrow R=2 \Omega }
According to Ohm's law
\mathrm{V+I_A R \Rightarrow 8=I_A \times 2 \Rightarrow I_A=4 \mathrm{~A} }
In circuit B, lower P-Njunction diode is reverse biased. Hence, no current will flow but upper diode is forward biased, so current can flow through it
\mathrm{V=I_B R \Rightarrow 8=I_B \times 4 \Rightarrow I_B=2 A }

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Rishi

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