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 = ${\displaystyle \frac{\text{V}}{\text{R}}=\frac{15}{12}}$

I = ${\displaystyle \frac{5}{4}}$

Electric energy spent in 6 Ω per minute

W = I² Rt

W = ${\displaystyle (\frac{5}{4}\times \frac{5}{4})\times 6\times \left(60\right)}$

= 562.5 J

when R₁ and R₂ are connected in parallel

${\displaystyle \frac{1}{{\text{R}}_3}=\frac{1}{6}+\frac{1}{4}}$

= ${\displaystyle \frac{5}{12}}$

∴ R₃ = ${\displaystyle \frac{12}{5}}$ Ω

∴ R₃ + r = ${\displaystyle \frac{12}{5}+2}$

= ${\displaystyle \frac{22}{5}}$ Ω

Total resistance of the circuit R = ${\displaystyle \frac{22}{5}}$ Ω

I = ${\displaystyle \frac{\text{V}}{\text{R}}}$

= ${\displaystyle \frac{15\times 5}{22}}$

= ${\displaystyle \frac{75}{22}}$ A

V₁ of parallel combination

V₁ = I R₃

= ${\displaystyle \frac{75}{22}\times \frac{12}{5}}$

= ${\displaystyle \frac{90}{11}}$ V

∴ I₁ through 6 Ω = ${\displaystyle \frac{{\text{V}}_1}{6}}$

= ${\displaystyle \frac{90}{11}\times \frac{1}{6}}$

= ${\displaystyle \frac{15}{11}}$ A

∴ Energy spent in 6 Ω resistor

W = I₁² R t

W = ${\displaystyle (\frac{15}{11}\times \frac{15}{11})\times 6\times \left(60\right)}$

= 669.1 J