What is the smallest number of 1Ω resistors needed such that when arranged in a certain arrangement involving only series and parallel connections, that the equivalent resistance is (7/6)Ω?

Emf =8V,

Internal resistance r=0.5Ω
R1 = R2 = 15 Ω

Equivalent capacitance of R1 and R2 is

R= 7.5 Ω

now,

emf = I (R+r)
8 = I (7.5+0.5)

8 = I 8
I = 1A

Now,

E= IR+ Ir
E = V + Ir
8 = V + 1x0.5

V= 7.5 volt

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