Norton’s Theorem

 

Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source I_N in parallel with a resistor R_N, where I_N is the short-circuit current through the terminals and R_N is the input or equivalent resistance at the terminals when the independent sources are turned off.

Figure 4.37 (a) Original circuit, (b) Norton equivalent circuit.

Thus, the circuit in Fig. 4.37(a) can be replaced by the one in Fig. 4.37(b). The proof of Norton’ s theorem will be given in the next section. For now, we are mainly concerned with how to get R_N and I_N. We find R_N in the same way we find R_Th. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is,

To find the Norton current I_N, we determine the short-circuit current flowing from terminal a to b in both circuits in Fig. 4.37. It is evident that the short-circuit current in Fig. 4.37(b) is IN. This must be the same short-circuit current from terminal a to b in Fig. 4.37(a), since the two circuits are equivalent. Thus,

Figure 4.38 Finding Norton current I_N.

shown in Fig. 4.38. Dependent and independent sources are treated the same way as in Thevenin’s theorem. Observe the close relationship between Norton’ s and Thevenin’s theorems: R_N = R_Th as in Eq. (4.9), and:

This is essentially source transformation. For this reason, source transformation is often called Thevenin-Norton transformation.

Since V_Th, I_N, and R_Th are related according to Eq. (4.11), to determine the Thevenin or Norton equivalent circuit requires that we find:

• The open-circuit voltage v_oc across terminals a and b.
• The short-circuit current i_sc at terminals a and b.
• The equivalent or input resistance R_in at terminals a and b when all independent sources are turned off.

We can calculate any two of the three using the method that takes the least effort and use them to get the third using Ohm’s law.

the open-circuit and short-circuit tests are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source.

Example:

Find the Norton equivalent circuit of the circuit at terminals a-b.

Solution:

We find R_N in the same way we find R_Th in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit in Figure: 

from which we find R_N. Thus:

To find IN, we short-circuit terminals a and b, as shown in Figure:

We ignore the 5-Ω resistor because it has been short-circuited. Applying mesh analysis, we obtain:

From these equations, we obtain:

Alternatively, we may determine I_N from V_Th ∕ R_Th. We obtain V_Th as the open-circuit voltage across terminals a and b in Figure:

 Using mesh analysis, we obtain: