JEE Main 2021 · 27 July · evening · Numerical answer · +4 / −0
For the circuit shown, the value of current at time t = 3.2 s will be _________ A.
[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]
Answer and solution
Answer 1
Solution
Physics sir · 3 steps
Read V(t) off the graph at 3.2 s, then notice the 5 V cell pushes back against it.
- Voltage at 3.2 s. Between t = 3 s and t = 4 s the voltage rises from 5 V to 10 V, so the slope is 5 V s⁻¹. At t = 3.2 s, V = 5 + 5 × 0.2 = 6 V.
- Direction of the cell. The positive terminals of V(t) and the 5 V cell both face the top wire, through R. So in the loop the two sources oppose each other, and the net emf is their difference.
I = (V(t) - 5)/R = (6 - 5)/1 = 1 A
- So. The current at t = 3.2 s is 1 A, flowing clockwise as marked.
The common slip: 11. Adding the two emfs instead of subtracting them gives (6 + 5)/1 = 11 A, not 1. The cell's positive terminal faces the source's positive terminal, so they oppose.
The idea to keep. In a single loop with two sources, check which terminals face each other. Like terminals facing means the emfs subtract. Then Ohm's law on the net emf gives the current.
Answer from the NTA final answer key. Solution by PrepWiser.