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.

Answer 1

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