Page 153 - Basic Electrical Engineering

Current of 2 A will be divided into two parts, one part going through the 8 Ω
resistor from B to A and the other part going through the 2 Ω resistor. Current
going through the 8 Ω resistor is calculated as

Now, we will consider the voltage source, keeping the current source open
circuited and find the current through the 8 Ω resistor. Using Ohm’s law, the

current through the 8 Ω resistor is calculated as

Figure 2.48

The combined effect of the two sources when superimposed is

2.8.2 Thevenin’s Theorem

Application of this theorem often comes useful when we want to determine
the current flowing through any branch or component of a network. We can

conveniently determine the current through any component when it is
required that the component be replaced. The use of kirchchoffs laws to

calculate the branch current for the changed value of a resistor becomes time
consuming as we have to repeat the calculations.