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Question
A battery of e.m.f 15 V and internal resistance 2 Ω is connected to two resistors of resistance 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.
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Solution
r = 2 Ω, R1 = 4 Ω, R2 = 6 Ω, V = 15 V
t = 60 S
As Resistance are in series
∴ Total resistance of circuit = R = 2 + 4 + 6 = 12 Ω
I in the circuit = `"V"/"R" = 15/12`
I = `5/4`
Electric energy spent in 6 Ω per minute
W = I2 Rt
W = `(5/4 xx 5/4) xx 6 xx (60)`
= 562.5 J
when R1 and R2 are connected in parallel
`1/"R"_3 = 1/6 + 1/4`
= `5/12`
∴ R3 = `12/5` Ω
∴ R3 + r = `12/5 + 2`
= `22/5` Ω
Total resistance of the circuit R = `22/5` Ω
I = `"V"/"R"`
= `(15 xx 5)/22`
= `75/22` A
V1 of parallel combination
V1 = I R3
= `75/22 xx 12/5`
= `90/11` V
∴ I1 through 6 Ω = `"V"_1/6`
= `90/11 xx 1/6`
= `15/11` A
∴ Energy spent in 6 Ω resistor
W = I12 R t
W = `(15/11 xx 15/11) xx 6 xx (60)`
= 669.1 J
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