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(B) In the given circuit,the top diode is reverse-biased because the positive terminal of the battery is connected to its cathode. Therefore,no curren

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

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(B) In the given circuit,the top diode is reverse-biased because the positive terminal of the battery is connected to its cathode. Therefore,no current flows through the top branch.The bottom diode is forward-biased because the positive terminal of the battery is connected to its anode. Thus,current flows through the bottom branch.The total resistance in the active branch is $R = 20 \, \Omega + 30 \, \Omega = 50 \, \Omega$.The voltage source is $V = 5 	ext{ V}$.Using Ohm's law,the current $i$ is given by $i = \frac{V}{R} = \frac{5}{50} 	ext{ A}$.

Find the current $i$ in the circuit.

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