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Question
A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
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Solution
(a) ε = 15 V
R = 6 + 3 = 9 ohm
r = 3 ohm
I = ?
I = ε / (R + r)
I = 15 / (9 + 3) = 15/12 = 1.25 A
(b) Current (calculated in (a) part) I = 1.25 A
External Resistance R = 6 + 3 = 9 ohm
V = IR = 1.25 × 9 = 11.25 V
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