Two identical batteries of emf = 13.6 V and internal resistance r = 0.100 are to

Question

Two identical batteries of emf = 13.6 V and internal resistance r = 0.100 are to be connected to an external resistance R, either in parallel (see Figure (a)) or in series (see Figure (b)). If R = 2.00r, what is the current i in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is i greater? If R=r/2.00, what is i in the external resistance in the (d)parallel arrangement and (e) series arrangement? (f) For
which arrangement is i greater now?

Explanation / Answer

given two identical batteries, E = 13.6 V

internal resistance, r = 0.1 ohm

external resistance = R = 2r = 0.2 ohm

a. for parallel connection

eazch battery gives a current i, where total current in the resistor is 2i

and E = ir + 2iR

13.6 = i(0.1 + 2*0.2) = i(0.1 + 0.4) = 0.5i

i = 27.2 A

hence current in the resistor = 2i = 54.4 A

b. for series arrangement

current in the circuit = i

then 2E = (R + 2r)i

2*13.6 = (0.2 + 2*0.1)i = 0.4i

i = 68 A

c. i is greater for series arrangement

d. for parallel connection

eazch battery gives a current i, where total current in the resistor is 2i

and E = ir + 2iR

13.6 = i(0.1 + 2*0.05) = i(0.1 + 0.1) = 0.2i

i = 68 A

hence current in the resistor = 2i = 136 A

e. for series arrangement

current in the circuit = i

then 2E = (R + 2r)i

2*13.6 = (0.05 + 2*0.1)i = 0.25i

i = 108.8 A

f. i is greater for parallel arrangement now

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