The current i in the network is: (in A) | vedclass

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(C) In the given circuit,the two diodes are connected in such a way that the top diode is forward-biased and the bottom diode is reverse-biased.Theref

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$0.3$

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(C) In the given circuit,the two diodes are connected in such a way that the top diode is forward-biased and the bottom diode is reverse-biased.Therefore,the bottom branch containing the reverse-biased diode acts as an open circuit.The circuit simplifies to a series combination of the $5 \ \Omega$ resistor (external),the $5 \ \Omega$ resistor (top middle),the $10 \ \Omega$ resistor (top right),and the $10 \ \Omega$ resistor (top left).The total resistance $R_{eq} = 5 \ \Omega + 5 \ \Omega + 10 \ \Omega + 10 \ \Omega = 30 \ \Omega$.The current $i$ is given by $i = \frac{V}{R_{eq}} = \frac{9 \ V}{30 \ \Omega} = 0.3 \ A$.

The current $i$ in the network is: (in $A$)

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