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(D) In the given circuit,both diodes are connected in parallel and are forward-biased by the $3 \, V$ battery.Each branch consists of a diode $(30 \,

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$0.03$

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(D) In the given circuit,both diodes are connected in parallel and are forward-biased by the $3 \, V$ battery.Each branch consists of a diode $(30 \, \Omega)$ in series with a $70 \, \Omega$ resistor. The total resistance of each branch is $R_{branch} = 30 \, \Omega + 70 \, \Omega = 100 \, \Omega$.Since there are two such identical branches in parallel,their equivalent resistance is $R_p = \frac{100 \, \Omega}{2} = 50 \, \Omega$.This parallel combination is in series with the $50 \, \Omega$ resistor and the $3 \, V$ battery.Total resistance of the circuit is $R_{total} = R_p + 50 \, \Omega = 50 \, \Omega + 50 \, \Omega = 100 \, \Omega$.The current through the $50 \, \Omega$ resistor is the total current of the circuit:$I = \frac{V}{R_{total}} = \frac{3 \, V}{100 \, \Omega} = 0.03 \, A$.

The circuit shown in the figure contains two diodes,each with a forward resistance of $30 \, \Omega$ and infinite backward resistance. If the battery is $3 \, V$,the current through the $50 \, \Omega$ resistance (in ampere) is

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