Question

A battery of e.m.f. 15 V and internal resistance 2 `Omega` is connected to two resistors of 4 ohm and 6 ohm joined.

(i) In series,

(ii) In parallel.

Find in each case the electrical energy spent per minute in 6 ohm resistor.

Answer 1

Given, emf= 15 V, internal resistance, r = 2 `Omega`

(i) Total resistance of given resistors in series, R_s = 4 + 6 = 10 `Omega`

Now, current through series combination on, I_s = `"emf"/(R_s + r) = 15/(10 + 2) = 15/12` = 1.25 A

In series, the same current will pass through each resistor

∴ Electrical energy spent per minute in the 6`Omega` resistor, H_s = I² (R) t = (1.25)² (6) (60)

= 562.5 J

(ii) Total resistance of given resistors in parallel, R_P = `(1/4 + 1/6) ^-1

= 2.4 Omega`

Current in parallel circuit, I = `15/(2.4 + 20) = 3.4 "A"`

In parallel, each resistor is connected across the same voltage, say V.

Then, V = e - Ir= 15 - (3.4 x 2) = 8.18 volt

∴ Electrical energy spent per minute in the 6Ω resistor, H_p = `V^2/R t = (8.18)^2/6 xx 60 `= 669.1 J