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Question

(A) When the positive terminal of the $2V$ battery is connected to point $A$,the upper diode is forward-biased,while the lower diode is reverse-biased

Vedclass · Accepted Answer

$0.2$

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(A) When the positive terminal of the $2V$ battery is connected to point $A$,the upper diode is forward-biased,while the lower diode is reverse-biased.In forward bias,the diode acts as a short circuit (resistance = $0 \Omega$).In reverse bias,the diode acts as an open circuit (resistance = $\infty \Omega$).Therefore,the current flows only through the upper branch containing the $10 \Omega$ resistor.The total resistance of the circuit is $R = 10 \Omega$.The current $I$ supplied by the battery is given by Ohm's law: $I = \frac{V}{R}$.Substituting the values: $I = \frac{2V}{10 \Omega} = 0.2A$.

$A$ $2V$ battery is connected across the points $A$ and $B$ as shown in the figure. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias,the current supplied by the battery when its positive terminal is connected to $A$ is........$A$

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