A p-n junction diode is reverse biased by a 12 V battery through a 2 kΩ resistor in series. The reverse current in the circuit is 5 μA. The voltage across the resistor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10\ \mathrm{mV}\)
In reverse bias the diode passes only a very small current, of the order of microampere. The same current flows through the series resistor, so the voltage across the resistor is very small.
\( I=5\ \mu\mathrm{A}=5\times10^{-6}\ \mathrm{A} \)
\( R=2\ \mathrm{k\Omega}=2\times10^{3}\ \Omega \)
\( \begin{aligned} V_R &= IR \\ &= (5\times10^{-6})\times(2\times10^{3}) \\ &= 1\times10^{-2}\ \mathrm{V}=10\ \mathrm{mV} \end{aligned} \)
From the above, \( V_R=10\ \mathrm{mV} \). Hence, almost all of the 12 V battery voltage appears across the reverse-biased diode.
NCERT (2026-27):
p-n junction diode under reverse bias – “For the diode in reverse bias, the current is very small (~μA) and almost remains constant with change in bias. It is called reverse saturation current.”