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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 -