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Electric Circuits
Tech 101.
Superposition and Thevenin’s Theorems
Supplement
Prepared by Mike Crompton. (Rev. 10 March 2009)
Thevenin’s Theorem.
There are rare occasions when the application of Ohm’s law and the laws of
series/parallel circuits make it extremely difficult or impossible to solve an electronic
circuit. For these rare cases a man called Thevenin developed a theorem that can simplify
a circuit to a single voltage (The Thevenin voltage VTH) and a single resistance (Thevenin
resistance RTH).
The theorem involves a series of steps like disconnecting a designated load resistance and
calculating the voltage across the points where the load was connected. This produces
VTH (to be used later). The supply voltage is then replaced with a short circuit and the
total resistance of the remaining resistors is then calculated giving us RTH (also to be used
later). Next the original supply voltage is replaced with VTH and RTH is placed in series
with it and the designated load resistor. From the point of view of the load resistor the
circuit is now reduced to a single resistance with a voltage supply. A very simple
example of the theorem is given below.
Fig 1 at right shows the circuit with RL
already disconnected. Notice that the
actual value of RL is not given as it does
not affect the simplification of the circuit.
Vsupply +
6V
R2
40
-
RL
Fig 1
Step 2 (See Fig 2) is to use the voltage
divider formula to determine the voltage
across R2.
VR2 = VSUPPLY / RTOTAL x R2
Vsupply +
VR2 = 6V/ 60Ω x 40 Ω
6V
VR2 = 4 Volts
R1 20
R2
40
-
Vth
RL
Fig 2
This will be VTH and will be used later.
Step 3 (See Fig 3) is to replace the supply
voltage with a short circuit and calculate
the resulting total resistance. You will see
that we have produced a simple circuit
with R1 and R2 actually in parallel.
Vsupply
replaced
with
short.
RTOTAL = R1 x R2 / R1 + R2
R1 20
R2
40
Fig 3
RTOTAL = 20 x 40 / 20 + 40
RTOTAL
R1 20
800/ 60 = 13.3Ω
This will be RTH and will be used with VTH in the next step.
2
Rth
RL
The final step (See Fig 4) is to combine the
two Thevenin values into one very simple
circuit with a single voltage and a single
resistance in series with the load resistance.
Rth
Vth
The benefit of Thevenin’s theorem is that any
complicated circuit can be reduced to a voltage
and series resistor feeding the load resistance.
RL
Fig 4
Superposition Theorem
There are occasions when a circuit may have two different supply voltages connected at
different points. There is usually one part of the circuit were current is flowing as a result
of both voltages and can be in the same or opposite directions. Another theorem is used
to simplify the circuit and determine what current is flowing where! ‘Superposition
Theorem’ does just that.
Fig 1 below right shows a simple circuit with two power supplies, both connected with
the same polarity but to different parts of the circuit.
It can be seen that current from both
supplies will be flowing through R2 and in
the same direction (electron flow from –ve
to +ve or ‘up’). These currents will add
together, and to determine how much from
each supply we must theoretically
eliminate one supply, calculate the current,
then eliminate the other supply and recalculate the current.
V1
supply
6V
R1 20
R2
40
-
R3 60
+
V2
+ supply
12V
-
Fig 1
Fig 2 shows supply 2 replaced by a short
circuit putting R2 and R3 in parallel. The
equivalent resistance will be,
REQ = R2 x R3 / R2 + R3
= 40 x 60 / 40 + 60
REQ = 24 Ω
V1
supply
6V
R1 20
R2
40
-
R3 60
+
IR3
Itotal
Fig 2
V2
replaced
with
short
circuit
IR2
The circuit now has R1 of 20 Ω and REQ of 24 Ω in series circuit with a 6V supply. The
voltage across the REQ (R2 and R3 in parallel) using the voltage divider formula will be,
VREQ = VTOTAL/ RTOTAL x REQ
VREQ = 6 / 44 x 24 = 3.27V
Total current can therefore be calculated as,
ITOTAL = VTOTAL/ RTOTAL
= 6V / 44 Ω
ITOTAL = .0136A or 136mA
3
This total current of 136.4 mA will now split between R2 and R3. We are, at this time,
only interested in the current through R2. It can be calculated using the voltage across the
two parallel resistors found earlier as 3.27V,
IR2 = VR2 / R2
IR2 = 3.27 / 40 = 0.08175A or 81.75mA
This current will flow ‘up’ through R2 and will be added to the current resulting from
Power supply #2 which we will now calculate.
Fig 3 shows Power supply #1 replaced with a
short circuit, placing R1 and R2 in parallel
for an REQ of,
REQ = R1 x R2 / R1 + R2
= 20 x 40 / 20 + 40
REQ = 13.3 Ω
V1
replaced
with
short
circuit
R1 20
R2
40
IR1
R3 60
IR2
Itotal
Fig 3
The circuit now has R3 of 60 Ω and REQ of 13.3 Ω in series circuit with a 12V supply.
The voltage across REQ (R1 and R2 in parallel) using the voltage divider formula will be,
V REQ = VTOTAL/ RTOTAL x REQ
REQ = 12 / 73.3 x 13.3 = 2.18V
Total current can therefore be calculated as,
ITOTAL = VTOTAL/ RTOTAL
= 12V / 73.3 Ω
ITOTAL = .1637A or 163.7mA
This total current of 163.7 mA will now split between R2 and R3. Once again we are
only interested in the current through R2. It can be calculated using the voltage across the
two parallel resistors found earlier as 2.18V,
IR2 = VR2 / R2
IR2 = 2.18 / 40 = 0.0545A or 54.5mA
This current will also flow ‘up’ through R2 and will be added to the current resulting
from Power supply #1 giving a total current flow through R2 of 81.75mA + 54.5mA =
136.25mA
If either of the two power supplies had the opposite polarity the calculations would be
the same BUT the currents would be subtracted from each other. In the case of supply
#2 being reversed, current flow through R2 would be in the same direction (up) but
would be 81.75mA - 54.5mA = 27.25mA.
Should supply #1 be reversed the same amount of current would flow through R2 BUT in
the opposite direction (down).
4
V2
+ supply
12V
-