A silicon diode and a resistor are connected in series with a 5.7 V battery so that the diode is forward biased. The current in the circuit is to be 25 mA. Take the voltage across the conducting diode as 0.7 V, its cut-in voltage. The resistance needed is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(200\ \Omega\)
When the silicon diode conducts, the voltage across it stays near its cut-in value of 0.7 V. The rest of the battery voltage appears across the resistor.
\( V_B=5.7\ \mathrm{V} \)
\( V_D=0.7\ \mathrm{V} \)
\( I=25\ \mathrm{mA}=0.025\ \mathrm{A} \)
The voltage across the resistor is
\( \begin{aligned} V_R &= V_B-V_D \\ &= 5.7-0.7 \\ &= 5.0\ \mathrm{V} \end{aligned} \)
The resistance needed is
\( \begin{aligned} R &= \frac{V_R}{I} \\ &= \frac{5.0}{0.025} \\ &= 200\ \Omega \end{aligned} \)
NCERT (2026-27):
p-n junction diode under reverse bias – “… This voltage is called the threshold voltage or cut-in voltage (~0.2V for germanium diode and ~0.7 V for silicon diode).”