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(B) In the given circuit,the diode in the upper branch is in reverse bias. Therefore,no current flows through the $20 \, \Omega$ resistor in that bran

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$\frac{5}{50}$

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(B) In the given circuit,the diode in the upper branch is in reverse bias. Therefore,no current flows through the $20 \, \Omega$ resistor in that branch.Assuming the forward bias resistance of the diode in the middle branch is zero,the total resistance of the circuit is the sum of the $20 \, \Omega$ resistor in series with the battery and the $30 \, \Omega$ resistor in the middle branch.Total resistance $R = 20 \, \Omega + 30 \, \Omega = 50 \, \Omega$.The current $I$ flowing in the circuit is given by Ohm's law: $I = \frac{V}{R} = \frac{5 \, V}{50 \, \Omega} = \frac{5}{50} \, A$.

What will be the current flowing in the circuit?

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