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Question
A battery of e.m.f. 12V and internal resistance 1.6 Ω is connected to two resistors of 4Ω and 6Ω connected in parallel. Calculate:
(i) the current drawn from the battery,
(ii) the power dissipated in each resistor,
(ii)the total power supplied by the battery.
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Solution
Given: E = 12V, r = 1.6 Ω, R1 = 4 Ω, R2 = 6 Ω
Equivalent resistance in parallel R = `("R"_1"R"_2)/("R"_1 + "R"_2) = (4 xx 6)/(4 + 6) = 2.4 Omega`
Total resistance of circuit = R = r = 2.4 + 1.6 = 4.0 Ω
(i) The current drawn from the battery
I = `"E"/("R" + "r") = 12/4.0 = 3.0` A
(ii) The voltage drop in internal resistance v = Ir = 3.0 × 1.6 = 4.8 V
Voltage across each resistor (connected in parallel) V = E - v = 12 - 4.8 = 7.2 V
Power dissipated in R1 = 4 Ω, P1 = `"V"^2/"R"_1 = (7.2)^2/4 = 12.96` W
Power dissipated in R2 = 6 Ω, P2 = `"V"^2/"R"_2 = (7.2)^2/6 = 8.64` W
(iii) Total power supplied by the battery P = El = 12 × 3 = 36 W.
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