For the given diagram of barrier potential of this diode is 0.5 V. The value of current in the circuit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{20}\,\mathrm{A} \)
All the elements are in one series loop, so the same current flows through each. The longer plate of each cell is the positive terminal, and it is on the left. The current leaves the battery towards the \( 90\,\Omega \) resistor and enters the diode at its p-side, so the diode is forward biased.
A forward-biased diode conducts only after the barrier potential of 0.5 V is overcome, so 0.5 V of the supply is used up across it and the rest drops across the resistor.
\( \displaystyle I = \frac{V - V_{B}}{R} = \frac{5 - 0.5}{90} = \frac{4.5}{90} = \frac{1}{20}\,\mathrm{A} \)