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(D) In the given circuit,the diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery. The diode $D_2$ is

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(D) In the given circuit,the diode $D_1$ is forward-biased because its p-side is connected to the positive terminal of the battery. The diode $D_2$ is reverse-biased because its p-side is connected to the negative terminal of the battery.Therefore,no current flows through the branch containing $D_2$.The circuit effectively consists of the battery $(5 \ V)$,the resistor $(20 \ \Omega)$,and the branch containing $D_1$ with a resistor $(30 \ \Omega)$ in series.The total resistance of the circuit is $R_{eq} = 20 \ \Omega + 30 \ \Omega = 50 \ \Omega$.Using Ohm's law,the current $I$ flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{5 \ V}{50 \ \Omega} = 0.1 \ A$.

If the internal resistance of the cell is negligible,then the current flowing through the circuit is: (in $A$)

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