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(B) In the given circuit,the battery of $5 	ext{ V}$ is connected in series with a $20 	ext{ } \Omega$ resistor and two parallel branches.The upper

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$\frac{5}{50} 	ext{ A}$

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(B) In the given circuit,the battery of $5 	ext{ V}$ is connected in series with a $20 	ext{ } \Omega$ resistor and two parallel branches.The upper branch contains a $20 	ext{ } \Omega$ resistor and a diode. The diode is connected such that its p-side is towards the right,making it reverse-biased with respect to the positive terminal of the battery.The middle branch contains a diode and a $30 	ext{ } \Omega$ resistor. The diode is connected such that its p-side is towards the left,making it forward-biased with respect to the positive terminal of the battery.Since the upper branch is reverse-biased,it acts as an open circuit (no current flows through it).The total resistance of the circuit is the sum of the $20 	ext{ } \Omega$ resistor in the main line and the $30 	ext{ } \Omega$ resistor in the active middle branch: $R_{eq} = 20 	ext{ } \Omega + 30 	ext{ } \Omega = 50 	ext{ } \Omega$.Using Ohm's law,the current $i$ is:$i = \frac{V}{R_{eq}} = \frac{5 	ext{ V}}{50 	ext{ } \Omega} = \frac{5}{50} 	ext{ A}$.

The current $i$ in the circuit will be:

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