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Question
Four cells, each of e.m.f. 1.5 V and internal resistance 2.0 ohms are connected in parallel. The battery of cells is connected to an external resistance of 2.5 ohms. Calculate:
(i) The total resistance of the circuit.
(ii) The current flowing in the external circuit.
(iii) The drop in potential across-the terminals of the cells.
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Solution
E.m.f. of cell (each) = 1.5 V. and also e.m.f. of all four cells = 1.5V, internal resistance of each cell = 2.0 ohm.
Total internal resistance of all four cells
`1/"r"^1 = 4/"r" = 4/2.0 = 2.0 "ohm"`
∴ r1 = `1/2` = 0.5 ohm
(i) ∴ Total resistance of circuit = R + r
= 2.5 + 0.5 = 3.0 ohm
(ii) Current flowing in the external circuit
I = `"E"/"r" = 1.5/2.5 = 0.6` Amp.
(iii) The drop in potential across the terminals of the cells
= `"I" xx "r"/4 = 0.6 xx 2/4 = 0.6 xx 0.5 = 0.30 "Volt."`
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