A) 1 A, 2 A
B) 2 A, 1 A
C) 4 A, 2 A
D) 2 A, 4 A
Correct Answer: C
Solution :
In circuit A, both (p-n) junction diode act as forward biasing. Hence, current flows in circuit A. Total resistance R is given by \[\frac{1}{R}=\frac{1}{4}+\frac{1}{4}\] or \[\frac{1}{R}=\frac{2}{4}\] or \[R=2\Omega \] According to Ohm s law \[V={{I}_{A}}R\] or \[8={{I}_{A}}\times 2\] or \[{{I}_{A}}=4A\] In circuit B, lower p-n-junction diode is reverse biased. Hence, no current will flow but upper diode is forward biased so current can flow through it \[V={{I}_{B}}R\] or \[8={{I}_{B}}\times 4\] or \[{{I}_{B}}=2A\]
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