JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :
- (a) 0.50 W
- (b) 0.072 W
- (c) 0.10 W
- (d) 0.125 W
Answer and solution
Answer (c) 0.10 W
Solution
Physics sir · 3 steps
Same current flows through r and R, so split the power in the same ratio as the voltage drops.
- Find the current. The terminal voltage of 2.5 V is the voltage across the external resistor. So I = P/V = 0.5/2.5 = 0.2 A.
- Voltage lost inside. The EMF is 3.0 V but only 2.5 V reaches the terminals. The other 3.0 - 2.5 = 0.5 V is dropped across the internal resistance r.
Pᵣ = (ε - V) I = 0.5 × 0.2 = 0.1 W
- So. The power dissipated in the internal resistance is 0.10 W, which is option (c). As a check, r = 0.5/0.2 = 2.5 Ω and I² r = 0.04 × 2.5 = 0.1 W.
The idea to keep. In a series loop the current is common, so each part's power is its voltage drop times that current. The drop across r is always EMF minus terminal voltage.
Answer from the NTA final answer key. Solution by PrepWiser.