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Two circuits made up of identical ideal emf devices (E- 1.64 V) and resistors (R-39.00) are shown in the figure below. What i
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Answer #1

(1) the potential difference

(a) for circuit 1

V_{1ab}=\epsilon

V_{1ab}=1.64V

(b) for circuit 2

V_{2ab}=2\epsilon

V_{2ab}=3.28V

(2)

(a) the current for circuit 1

I_{1ab}=V_{1ab}/R

I_{1ab}=1.64/39.0

I_{1ab}=0.04A

(b) for circuit 2

I_{2ab}=V_{2ab}/R

I_{2ab}=3.28/39.0

I_{2ab}=0.08A

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