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Question

For the circuit given below, evaluate the current (I) passing through 2 Ω.


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Solution

According to the kirchhoff’s voltage law (KVL) - the sum of all voltages around a closed loop in any circuit must be equal to zero.
Let us assume the direction of current (I) is clockwise.

Using Kirchoff's law's sign conventions as shown below:
From loop CABDC
4(I×2)6(I×4)=0
4(I×2)6(I×4)=0
10(6I)=0
6I=10
I=106
I=53
I=1.66A
As the direction current comes out to be negative.So, the direction of current will be opposite to the assumed direction.
Hence, 1.66 A current will flow in the 2 Ω resistance from B to A.

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