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Question

In the circuit shown E1=4.0V,R1=2Ω,E2=6.0V,R2=4Ω and R3=2Ω. The current I1 is
697267_4c21c58af5d846ba9b0501cda9aa5f34.png

A
1.6A
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B
1.8A
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C
1.25A
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D
1.0A
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Solution

The correct option is B 1.8A
Apply KVL in ABCD
42I14I2+6=02I1+4I2=10I1+2I2=5 (1)
$
Apply KVL in EDCF
42I1+2(I2I1)=042I1+2I22I1=04I12I2=4 (2)
Add (1) and (2)
5I1=9I1=9A5=1.8A

900685_697267_ans_61e9306a94c54121bf6ba749309e295b.png

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