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Forward Bias

Current flowi...

Question

Current flowing in each of the following circuits A and B respectively are the forward resistance of diodes

1 A and 2 A

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2 A and 1 A

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4 A and 2 A

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2 A and 4 A

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Solution

The correct option is A 4 A and 2 A
In circuit A, both the diodes are forward biased. Hence they ideally act as shorts. Thus equivalent resistance of the combination is $2 Ω$ .
Thus current in circuit A is $\frac{8 V}{2 Ω} = 4 A$ .

In circuit B, one diode is forward biased and other is reverse biased. Thus one is shorted(forward) and other is open(reverse). Hence effectively there is only one resistance in the circuit( $4 Ω$ ).
Thus current in circuit B is $\frac{8 V}{4 Ω} = 2 A .$

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