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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.
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Solution
Given, emf= 15 V, internal resistance, r = 2 `Omega`
(i) Total resistance of given resistors in series, Rs = 4 + 6 = 10 `Omega`
Now, current through series combination on, Is = `"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, Hs = I2 (R) t = (1.25)2 (6) (60)
= 562.5 J
(ii) Total resistance of given resistors in parallel, RP = `(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, Hp = `V^2/R t = (8.18)^2/6 xx 60 `= 669.1 J
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