## Thursday, November 13, 2008

### AIEEE 2008 Question Involving Kirchoff’s Laws

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:

A 5 V battery with internal resistance 2 Ω and a 2 V battery with internal resistance 1 Ω are connected to a 10 Ω resistor as shown in the figure. The current in the 10 Ω resistor is

(a) 0.27 A, P1 to P2

(b) 0.27 A, P2 to P1

(c) 0.03 A, P1 to P2

(d) 0.03 A, P2 to P1

The circuit is redrawn, indicating the currents in the three branches. We have marked the directions of the currents I1 and I2 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. There is nothing to be worried about such situations since you will obtain a 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 ABP2P1 and P1P2CD we have respectively,

5 = 2×I1 + 10×(I1 I2) and

2 = 1×I2 10×(I1 I2)

The above equations can be rewritten as

12 I1 – 10 I2 = 5 and

10 I1 + 11 I2 = 2

These equations can be easily solved to give I1 = 2.34 A (nearly) and I2 = 2.31 A (nearly) so that the current (I1 I2) = 0.03 A.

Since the currents I1 and I2 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 P2 to P1 [Option (4)].

[Suppose we had marked the current I2 as flowing in the opposite direction. The current flowing in the 10 Ω resistor will then be (I1 + I2). We will then obtain I1 = 2.34 A and I2 = – 2.31 A, the negative sign indicating that the real direction of I2 is opposite to what we marked. The current flowing through the 10 Ω resistor will again be obtained correctly as (I1 + I2) = 2.34 A– 2.31 A = 0.03 A].