Two identical batteries of emf ? = 10.8 V and internal resistance r = 0.242 ? ar

Question

Two identical batteries of emf ? = 10.8 V and internal resistance r = 0.242 ? are to be connected to an external resistance R, either in parallel (see Figure (a)) or in series (see Figure (b)). If R = 0.260 ?, what is the current i (in A) in the external resistance in the (a) parallel and (b) in the series arrangements?

fig27_43.gif

fig27_44.gif

Two identical batteries of emf ? = 10.8 V and internal resistance r = 0.242 ? are to be connected to an external resistance R, either in parallel (see Figure (a)) or in series (see Figure (b)). If R = 0.260 ?, what is the current i (in A) in the external resistance in the (a) parallel and (b) in the series arrangements?

Explanation / Answer

when in series:

I=21.6/(0.242+0.26)=43.02 A

when in parallel

I=21.6*0.242/(0.242+0.26)=10.4 A

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