In the given figure, a diode D is connected to an external resistance $R = 100 \, \Omega$ and an e.m.f of $3.5

Question

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

Accepted Answer

1. **Identify the Given Data:** E.m.f of the battery, $V = 3.5	ext{ V}$; Barrier potential of the diode, $V_d = 0.5	ext{ V}$; External resistance, $R = 100 \, \Omega$.
2. **Calculate Net Voltage:** Assuming the diode is forward-biased (which allows current to flow), it drops a voltage equal to its barrier potential. The net effective voltage driving the current through the external resistor is $V_{net} = V - V_d = 3.5	ext{ V} - 0.5	ext{ V} = 3.0	ext{ V}$.
3. **Calculate Current:** Using Ohm's law, the current in the circuit is $I = \frac{V_{net}}{R} = \frac{3.0	ext{ V}}{100 \, \Omega} = 0.03	ext{ A} = 30	ext{ mA}$.

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