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Question
A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistance 3 ohm and 6 ohm in series. Find:
(i) The current through the battery,
(ii) The p.d. between the terminals of the battery.
Solution
Given, emf, e = 15 v
Internal resistance, r = 3 `Omega`
Resistance of given two resistors in series, R, = 3 + 6 = 9 `Omega`
(i) Current = `"emf"/"Total resistance of circuit" = 15/(9 + 3) = 15/12` = 1.25 A
(ii) Potential difference= emf - voltage drop= e - (Ir)
Now, voltage drop= current x internal resistance= 1.25 x 3 = 3.75
.·. p.d. = 15 - 3.75= 11.25V
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