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].