For the circuit shown in the figure, the current I will be
(1) 0.5 A
(2) 0.75 A
(3) 1 A
(4) 1.5 A
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Ans: (3) 1 A
Sol.
By KVL in a closed loop ABCDA,
VA – I × 4 – I × 1 + 4 – I × 1 + 2 = VA
–6I + 6 = 0
I = 1 A
I=(2+4)÷(4+1+1)
I=6V÷6omega
I=1A
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