Two identical cells of emf 1.5 V each are connected in series across a 10 Ω resistor. An ideal voltmeter across the 10 Ω resistor reads 1.5 V. Find the internal resistance of each cell in Ω.
Physics · Current electricity · electric instruments
JEE Mains 11 April 2023 Shift 2 Evening Shift

Question

Two identical cells of emf 1.5 V each are connected in series across a 10 Ω resistor. An ideal voltmeter across the 10 Ω resistor reads 1.5 V. Find the internal resistance of each cell in Ω.

G1 Physics · Accepted Answer

Total emf = 3.0 V and total internal resistance = 2r. The voltmeter reading across 10 Ω gives current I = 1.5/10 = 0.15 A. Also I = 3/(10+2r). 0.15(10+2r)=3 1.5+0.3r=3 r=5 Ω. Final value: 5