Question

A battery of e.m.f. 15 V and internal resistance 2 Ω   is connected to two resistors of resistances 4 ohm and 6 ohm joined (a) in series. Find in each case the electrical energy spent per minute in 6 Ω resistor.

Answer 1

Given , E.m.f of battery , V = 15 V

Internal resistance of battery , R_B = 2 Ω

Resistance given in circuit , R₁ = 4 Ω

R₂ = 6 Ω

(i) When resistors are connected in series ,

Equivalent resistance , R = R_B + R₁ + R₂ = 12 Ω

Current in the circuit , I = `15/12` = 1.25 A

Now voltage across resistor R₂ , V₂ = IR = 1.25 × 6

V₂ = 7.50 V

Time , t = 1 min = 60 sec

Energy across R₂ , E = `("V"^2"t")/"R" = ((7.5)^2 xx 60)/6`

E = 562.5 J