The current I in the circuit shown below is: | vedclass

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(B) The circuit consists of four parallel branches connected to a $10 	ext{V}$ $DC$ source.Each branch contains a resistor and a diode.Analyzing the

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$\frac{15}{2} 	ext{A}$

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(B) The circuit consists of four parallel branches connected to a $10 	ext{V}$ $DC$ source.Each branch contains a resistor and a diode.Analyzing the polarity of the diodes with respect to the $10 	ext{V}$ battery:$1$. The top branch ($4 \Omega$ resistor) has the diode in forward-biased condition.$2$. The second branch ($3 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).$3$. The third branch ($2 \Omega$ resistor) has the diode in forward-biased condition.$4$. The bottom branch ($5 \Omega$ resistor) has the diode in reverse-biased condition (it acts as an open circuit).Only the branches with $4 \Omega$ and $2 \Omega$ resistors are active.These two resistors are in parallel,so the equivalent resistance $R_{	ext{eq}}$ is:$R_{	ext{eq}} = \frac{4 	imes 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \Omega$.The total current $I$ drawn from the battery is:$I = \frac{V}{R_{	ext{eq}}} = \frac{10}{4/3} = \frac{30}{4} = 7.5 	ext{A} = \frac{15}{2} 	ext{A}$.Thus,the correct option is $B$.

The current $I$ in the circuit shown below is: (All diodes are ideal and identical)

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