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Question
Two cells each having an e.m.f. of 2 V and an internal resistance of 2Ω are
connected (a) In series, and ( b) In para 1le l as shown in fig. . What is the
current flowing through the cir cu it in each_ case?
Solution
(a) Given, emf (e) of the battery= 2 x 2 = 4V,
internal resistance r = 2 x 2= 4Ω
Resistance in the circuit, R = 4 Ω
Total resistance of the given series circuit = 4 + 4 = 8Ω
Current flowing through the circuit, I = `"e"/("R" + "r") = 4/8 =0.5 "A"`
(b) Given, emf of battery ( e) = emf of each cell in parallel = 2V,
Total internal resistance `1/"r" = 1/2 + 1/2 = 1Ω`
Resistance connected in the circuit ,R = 4Ω
Total resistance of the circuit= 1 + 4 = 5 Ω
Current flowing through the circuit, I= `"e"/("R" + "r") = 2/5 = 0.4 "A"`
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