Thevenin and Norton Theorems
33.1 Introduction
Many of the networks analysed in chapters 30, 31, & 32 using Kirchhoff’s laws, mesh current analysis, nodal analysis, and the superposition theorem can be analysed more quickly and easily by using Thevenin and Norton theorems.[770-779]
Thevenin’s Theorem
This states: “The current which flows in any branch of a network is the same as that which would flow in the branch if it were connected across a source of electrical energy, the emf of which is equal to the potential difference which would appear across the branch if it were open circuited, and the internal impedance of which is equal to the impedance which appears across the open-circuited branch terminals when all sources are replaced by their internal impedances.”[780-791]
The theorem applies to any linear active network. The term “active” means it contains a source or sources of emf.[792, 793] The term “linear” means the measured values of circuit components are independent of the direction and magnitude of the current flowing through them.[794-796]
This theorem simply means that a complicated network with terminals A-B can be replaced by a simple source E in series with an impedance z.[798-804] E is the open circuit voltage measured at terminals A-B, and z is the equivalent impedance of the network at terminals A-B when all internal sources of e.m.f are made zero.[812-815]
Procedure for finding the Equivalent Circuit
The following procedure can be used when determining the current flowing in a branch:[821-824]
- Remove the load/target impedance ZL (break).[825-827]
- Determine the open circuit voltage E across the break.[828-830]
- Remove each source of emf and replace it by its internal impedance (if it has zero internal impedance replace it by a short circuit) and then determine the internal impedance looking in at the break.[831-836]
- Determine the current for the Thevenin equivalent circuit using:[837-840]
I = E / (Z + ZL)[851, 852]
38.4 Norton’s Theorem
In a Thevenin circuit, the source of electrical energy is represented by an EMF in series with an impedance.[1028] It can also be represented by a current source in parallel with an impedance; these two circuits are electrically equivalent.[1029, 1030]
Norton’s Theorem states: “The current that flows in any branch of a network is the same as that which would flow in the branch if it were connected across a source of electrical energy, the short-circuit current of which is equal to the current that would flow in a short-circuit across the branch, and the internal impedance of which is equal to the impedance which appears across the open-circuited branch terminals.”[1033-1049]
This means that a network containing voltage sources and impedances can be replaced by a current source in parallel with an impedance.[1050-1057]
Method for finding the Norton Equivalent Circuit
- Short circuit terminals A-B.[1072]
- Determine the short-circuit current ISC.[1073]
- Remove each source of emf and replace it with a short circuit or with an internal resistance. (If current sources exist, replace them with an open circuit), then determine the impedance Z ‘looking in’ at a break between A and B.[1074-1078]
- Determine the value of current flowing in impedance from the equation of the Norton equivalent network:[1080]
IL = (Z / (ZL + Z)) × ISC[1081]
Worked Problems
Problem 15 (Source Transformations)
Convert the circuit left of A-B to a Norton circuit, then back to a Thevenin circuit.[1240, 1241]
Given: E1 = 12V, Z1 = 3Ω; E2 = 24V, Z2 = 2Ω.[1246, 1249, 1250]
Norton Conversion:
ISC2 = E2 / 2Ω = 12A[1252]
Itotal = 4A + 12A = 16A[1252]
ZN = (3 × 2) / (3 + 2) = 1.2Ω[1258]
Thevenin Conversion: Multiply current by resistance to get voltage.
Z = 1.2Ω[1266]
Current in load ZL = (1.8 + j4)Ω:[1266]
Problem 16 (Successive Transformations)
Use successive Thevenin and Norton conversions to find a Thevenin equivalent for terminals A-B.[1270, 1271, 1303, 1304]
Given Source 1: 5V, 1kΩ. Source 2: 10V, 4kΩ.[1274, 1276, 1305, 1310]
ISC2 = 10V / 4000Ω = 2.5mA[1279, 1313]
ISC_total = 5mA + 2.5mA = 7.5mA[1281, 1315]
Parallel Resistance:
Converting back to Thevenin:
IL = 6V / (2.8kΩ + 0.2kΩ – j4kΩ) = 6 / (3k – j4k)
IL = 0.8∠53.13° mA[1299, 1300, 1332, 1333]