JEE Main 2021 · 25 February · morning · Single correct · +4 / −1
A 5V battery is connected across the points X and Y. Assume D₁ and D₂ to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.
- (a) ∼ 0.5 A
- (b) ∼ 1.5 A
- (c) ∼ 0.43 A
- (d) ∼ 0.86 A
Answer and solution
Answer (c) ∼ 0.43 A
Solution
Physics sir · 4 steps
With X positive, current wants to flow left to right. Check which diode that direction turns on.
- Find the direction. X joins the left wire and Y joins the right wire. With the positive terminal at X, conventional current tries to go from left to right through each branch.
- Check each diode. D₁ points to the right, so it is forward biased and conducts. D₂ points to the left, so it is reverse biased and the 5 Ω branch carries no current.
- Silicon drop. A conducting silicon diode takes about 0.7 V. The remaining voltage falls across the 10 Ω resistor.
I = (5 - 0.7)/10 = 4.3/10 = 0.43 A
- So. Only the D₁ branch conducts, and the battery supplies about 0.43 A, which is option (c).
The idea to keep. In diode circuits, first decide which diodes are forward biased for the given polarity. Remove the reverse-biased branches, and for silicon subtract 0.7 V before using Ohm's law.
Answer from the NTA final answer key. Solution by PrepWiser.