Page 178 - Basic Electrical Engineering

or,

Similarly from loop N T R S,

+24 − 4I − 4I − 4(I + I ) = 0
2
2
1
2
or, 8 I = 24 − 4(I + I )
2
1
2
Substituting the value of (I + I )
1
2

According to Thevenin’s,

Thus, we get the same value of current through the load resistance. In circuit

solutions, experience will tell you as to which theorem is more suitable for
which kind of circuit solution.

Example 2.30 Apply Norton’s theorem to determine the current flowing
through the resistance of 6 Ω connected across the terminals. A and B. Also
calculate the potential of point A. What will be the current through the 6 Ω

resistor across AK. Solve this problem using Thevenin’s theorem also.