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(B) The potential difference across the resistance $R$ is given by subtracting the barrier potential of the diode from the total $e.m.f.$ of the circu

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$30$

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(B) The potential difference across the resistance $R$ is given by subtracting the barrier potential of the diode from the total $e.m.f.$ of the circuit:$V_R = E - V_{	ext{barrier}} = 3.5 \,V - 0.5 \,V = 3.0 \,V$According to Ohm's law,the current $I$ in the circuit is:$I = \frac{V_R}{R} = \frac{3.0 \,V}{100 \,\Omega} = 0.03 \,A$To convert the current into milliamperes $(mA)$:$I = 0.03 \,A 	imes 1000 \,mA/A = 30 \,mA$Therefore,the current in the circuit is $30 \,mA$.

In the given figure,a diode $D$ is connected to an external resistance $R = 100 \,\Omega$ and an $e.m.f.$ of $3.5 \,V$. If the barrier potential developed across the diode is $0.5 \,V$,the current in the circuit will be ........ $mA$.

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