The current in the circuit will be:

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(B) In the given circuit,the diode in the upper branch is reverse-biased because its p-terminal is connected to the negative side of the battery throu

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

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(B) In the given circuit,the diode in the upper branch is reverse-biased because its p-terminal is connected to the negative side of the battery through the circuit path,while the diode in the middle branch is forward-biased.Since the upper branch is open (due to the reverse-biased diode),the current flows only through the middle branch and the bottom branch.The total resistance in the active path is the sum of the resistor in the middle branch $(30 \, \Omega)$ and the resistor in the bottom branch $(20 \, \Omega)$.Using Ohm's law,$I = \frac{V}{R_{total}} = \frac{5 \, V}{20 \, \Omega + 30 \, \Omega} = \frac{5}{50} \, A$.

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