Question

14(d) Two cells, one having an emf of 2.0V and an internal resistance of 0.4Ω and the other having an emf of 2.0V and an internal resistance of 0.1Ω, are connected in parallel. The combination is then connected in series with a 5Ω resistor. (i) Draw a circuit diagram of the arrangement. (ii) Calculate the current through the 5Ω resistor.

Accepted Answer

(i) Circuit: two cells (V1=2.0V, r1=0.4Ω and V2=2.0V, r2=0.1Ω) connected in parallel, in series with a 5Ω resistor.
(ii) Since both cells have the same emf (2V), the total voltage across the parallel combination = 2V. The total internal resistance of the parallel combination is given by 1/r = 1/r1 + 1/r2 = 1/0.4+1/0.1 = 2.5+10 = 12.5, so r = 1/12.5 = 0.08Ω.
Using V = I(R+r): I = V/(R+r) = 2/(5+0.08) = 0.39 A.

WAEC Physics 2011 Theory — Question 33

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