Question : Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of 20 Ω. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :
a) 2.5 Ω
b) 4 Ω
c) 5 Ω
d) 10 Ω
[JEE Mains 2022]
Solution :
Let
r1=r2 = r (Internal resistance of each cell)
ε1 = ε2 = 1.5 V
R1 = R2 = 20 Ω
Terminal Potential difference OR Potential drop across the external resistance OR Reading in Voltmeter (V) = 1.2 V
Equivalent Emf of the cells (εeq)
= (ε1r2+ε2r1)/(r1+r2)
= (εr+εr)/(r+r)
= 2εr/2r
εeq = ε
Equivalent external resistance (Req)
= R1R2/(R1+R2)
= R²/2R
= R/2
= 20/2
Req = 10 Ω
= (εr+εr)/(r+r)
= 2εr/2r
εeq = ε
Equivalent external resistance (Req)
= R1R2/(R1+R2)
= R²/2R
= R/2
= 20/2
Req = 10 Ω
Equivalent internal resistance (req)
req = r1r2/(r1+r2)
req = r²/(r+r)
req = r²/2r
req=r/2
Now,
req=[(εeq/V)-1]Req
r/2 = [(ε/V)-1]Req
r/2 = [(1.5/1.2)-1]×10
r/2 = (1.25-1)×10
r/2 = 0.25×10
r/2 = 2.5
r = 2.5×2
req = r1r2/(r1+r2)
req = r²/(r+r)
req = r²/2r
req=r/2
Now,
req=[(εeq/V)-1]Req
r/2 = [(ε/V)-1]Req
r/2 = [(1.5/1.2)-1]×10
r/2 = (1.25-1)×10
r/2 = 0.25×10
r/2 = 2.5
r = 2.5×2
r = 5 Ω
Hence, c) 5Ω would be the correct option.
