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Question
A cell supplies a current of 1.2 A through two 2Ω resistors connected in parallel. When the resistors are connected in series, if supplies a current of 0.4 A. Calculate the internal resistance and e.m.f of the cell.
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Solution
(i) The effective resistance of a circuit with two 20-ohm resistors linked in parallel is R is `1/"R" = 1/2 + 1/2`
∴ R = 1Ω
'r' stands for the internal resistance of the cell that is in series with the circuit.
∴ V = IR
or V = 1.2 (1 + r) ...(i)
Connected in series, the effective resistance of the two resistors and the cell's internal resistance, denoted as 'r'
R = 2 + 2 + r
= (4 + r) Ω ...(ii)
V = IR
= 0.4 (4 + R)
(i) = (ii)
1.2 (1 + r) = 0.4 (4 + r)
1.2 + 1.2 r = 1.6 + 0.4 r
or (1.2 - 0.4) r = (1.6 - 1.2)
0.8 r = 0.4
r = `0.4/0.8`
= `1/2`
= 0.5 Ω
put r = 0.5 in (i)
(ii) e.m.f. of cell E = V = 1.2 (1 + 0.5) = 1.8 V
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