Two identical batteries of emf 12.0 V and internal resistance r = 0.200 are to be connected to an external resistance R, either in parallel or in series. If R =2.00r, what is the current i in the external resistance in the

a. parallel arrangements
b. series arrangements
c. For which arrangement is i greater If R= r/2.00, what is i in the external resistance?

a. i = 20A

b. i = 40A

c. series arrangement

Explanation:

a. In parallel, total emf E = 12.0v

r = 0.2Ω, R = 2.00r, hence, R = 2.00 x 0.2 = 0.4 Ω,

Recall; E = i (R + r)

12 = i (0.4 + 0.2)

12 = i (0.6)

i = 12/0.6

i = 20 A (for parallel)

b. In series, total emf E = E1 + E2 = 12.0 + 12.0 = 24.0v

r = 0.2Ω, R = 0.4Ω, i = ?

Recall; E = i (R + r)

24 = i (0.4 + 0.2)

i = 24/0.6

i = 40 A

c.  If R= r/2.00 in parallel arrangement

R = 0.2/2 = 0.1Ω

E = i (R + r)

12 = i (0.1 + 0.2)

12 = i (0.3)

i = 12/0.3

i = 40A

If R= r/2.00 in series arrangement

R = 0.2/2 = 0.1Ω

E = i (R + r)

24 = i (0.1 + 0.2)

12 = i (0.3)

i = 24/0.3

i = 80A

Hence, i is greater in series arrangement than parallel arrangement

Explanation:

d is the answer i came up with

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