Occasionally you will find questions requiring the application of Kirchoff’s laws. Here is a question that appeared in AIEEE 2008 question paper which I give here for clearing your doubts regarding the direction of the current you will have to mark in closed loops in direct current networks:

The current in the 10 Ω resistor is

(a) 0.27 A, P_{1} to P_{2}

(b) 0.27 A, P_{2} to P_{1}

(c) 0.03 A, P_{1} to P_{2}

_{2}to P

_{1}

The circuit is redrawn, indicating the currents in the three branches. We have marked the directions of the currents I_{1} and I_{2} in the directions we normally expect the 5 V and the 2 V batteries to drive their currents. [Note that there can be situations in which the direction we mark is wrong.

*negative*current as the answer and you will understand that the real direction is opposite to what you have marked in the circuit].

Applying Kirchoff’s voltage law (loop law) to the loops ABP_{2}P_{1} and P_{1}P_{2}CD we have respectively,

5 = 2×I_{1} + 10×(I_{1 }– I_{2}) and

2 = 1×I_{2 }–10×(I_{1 }– I_{2})

The above equations can be rewritten as

12 I_{1 }– 10 I_{2} = 5 and

–10 I_{1} + 11 I_{2} = 2

These equations can be easily solved to give I_{1} = 2.34 A (nearly) and I_{2} = 2.31 A (nearly) so that the current (I_{1 }– I_{2}) = 0.03 A.

Since the currents I_{1} and I_{2} are obtained as positive, the directions we marked are correct and the current flowing in the 10 Ω resistor is 0.03 A, flowing from P_{2} to P_{1} [Option (4)]. ** **

[Suppose we had marked the current I_{2} as flowing in the opposite direction. The current flowing in the 10 Ω resistor will then be (I_{1 }+ I_{2}). We will then obtain I_{1} = 2.34 A and I_{2} = – 2.31 A, the negative sign indicating that the real direction of I_{2} is *opposite to* what we marked. The current flowing through the 10 Ω resistor will again be obtained correctly as (I_{1 }+ I_{2}) = 2.34 A– 2.31 A = 0.03 A].

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