05 Thievenin Equivalent Circuit FINAL
THEVENIN EQUIVALENT CIRCUIT | C ONCEPT OVERVIEW
The topic of SOURCE EQUIVALENTS can be referenced on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. THEVENIN’S THEOREM states that a linear two-terminal network can be replaced with an equivalent circuit of a single voltage source and a series resistor. R TH
a
a V TH
b
b
• VTh is equal to the Voltage across terminals a & b when a & b are open.
• R Th is the resistance between a & b when the independent source is deactivated (voltage supply shorted or current source removed).
The Thevenin equivalent resistance is equal to the open-circuit voltage, VOC , divided by the short-circuited current, ISC . The Thevenin equivalent is:
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R eq =
Voc Isc
The formulas for the THEVENIN EQUIVALETN can be referenced under the topic of SOURCE EQUIVALENTS on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. The open circuit voltage Voc is Va -Vb , and the short circuit is the current ISC , from a to b. When a circuit is SHORT-CIRCUITED the current is said to be INFINITE.
STEPS FOR SOLVING THEVENIN EQUIVALENT CIRCUIT PROBLEMS: 1. Open the load resistor and remove the load resistance, R , from the circuit. L
2. Open current sources, and short voltage sources, by replacing the voltage source by its INTERNAL RESISTANCE.
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3. Calculate the equivalent resistance of the circuit looking from the load terminal towards the circuit. This is called the Thevenin’s Equivalent Resistance, R TH . 4. Calculate the SOURCE VOLTAGE, VS , by reconnecting the voltage sources, and finding the difference between the two voltage sources. 5. Calculate the SOURCE CURRENT, IS , using Ohm’s Law after finding the source voltage, VS , and the equivalent resistance with the source voltage connected, R S . 6. Calculate the OPEN CIRCUIT VOLTAGE, the Thevenin Voltage VTH , across each resistor, making sure to account for which voltage source is supplying the voltage to that resistor. This is done using Ohm’s Law. We can then calculate the voltage using the previously calculated source current in Step (5) and given resistance of the resistor. 7. Redraw the circuit with the calculated open circuit Voltage, VTH , in Step (6) as the voltage source and the calculated open circuit resistance R TH in step (3) as a series resistance. Connect the load resistor, R L , which we had removed in Step (1). This is the Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed by Thevenin’s Theorem. 8. We can now calculate the total current flowing through the circuit using Ohm’s Law expressed as: I T = 3|PREPINEER.COM
VTH . R TH + R L
psst…don’t forget to take notesÎ
CONCEPT EXAMPLE:
What is the Thevenin equivalent at terminal a-b for the circuit below? Calculate the total current flowing through the circuit.
SOLUTION:
1. Remove the load resistance, R L , from the circuit. In this problem, it is the 40Ω . 2. Open the current sources and short the voltages sources. The circuit would be drawn like the circuit below:
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3. Calculate the equivalent resistance looking form the load terminal towards the circuit. The 10Ω resistor is in parallel with the 20Ω resistor, so the equivalent resistance is calculated as:
R TH =
R 1 × R 2 20 × 10 = = 6.67Ω R 1 +R 2 20 + 10
4. Calculate the source voltage, VS , by reconnecting the two voltage sources in the circuit and taking the difference.
VS = V1 − V2 = 20V − 10V = 10V 5. Now, calculate the source current flowing in the circuit using Ohm’s Law. Plug inthe previously calculated source voltage and the equivalent resistance with the source voltage connected:
IS =
VS 20v-10v = = 0.33 amps R S 20!+10!
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6. Calculate the OPEN CIRCUIT VOLTAGE, the Thevenin Voltage, VTH , across each resistor using Ohm’s Law. Be sure to account for which voltage source is supplying the voltage to which resistor. Plugging in the previously calculated source current in Step (5) and given resistance of the resistor, calculate the voltage. The resistors are in series, so the current will be in the same in each of the resistors. Calculating the voltage drop across the 20Ω resistor with the 20 V voltage source:
(
)
VAB = V2 – R 20Ω x IS = VTH
(
)
VAB = 20 – 20" x 0.33 A = 13.33 volts Calculating the voltage drop across the 20Ω resistor with the 20 V voltage source:
(
)
VAB = V1 – R 10Ω x IS = VTH
(
)
VAB = 10 + 10" x 0.33amps = 13.33 volts
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The Thevenin Voltage is the same across each resistor:
VTH = 13.33Volts The Thevenin Equivalent Voltage is the same through each resistor. Just remember that since terminal a to terminal b is short circuited, the voltage cannot flow through it Therefore the calculation must use the voltage source that is directly connected to the resistor. 7. Redraw the circuit with the calculated open circuit Voltage, VTH , from Step (6) as the voltage Source. Then, connect the calculated open circuit resistance R TH from step (3) and the the load resistor,
R L , which we had removed in Step (1) in series. This is the
Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed by Thevenin’s Theorem.
8. We can now calculate the total current flowing through the circuit using Ohm’s Law expressed as: 7|PREPINEER.COM
psst…don’t forget to take notesÎ
IT =
VTH R TH + R L
This formula is not supplied in the Reference Handbook.
IT =
VTH 13.33v = = 0.286 amps R TH 6.67!+40!
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psst…don’t forget to take notesÎ
The topic of SOURCE EQUIVALENTS can be referenced on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. THEVENIN’S THEOREM states that a linear two-terminal network can be replaced with an equivalent circuit of a single voltage source and a series resistor. R TH
a
a V TH
b
b
• VTh is equal to the Voltage across terminals a & b when a & b are open.
• R Th is the resistance between a & b when the independent source is deactivated (voltage supply shorted or current source removed).
The Thevenin equivalent resistance is equal to the open-circuit voltage, VOC , divided by the short-circuited current, ISC . The Thevenin equivalent is:
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psst…don’t forget to take notesÎ
R eq =
Voc Isc
The formulas for the THEVENIN EQUIVALETN can be referenced under the topic of SOURCE EQUIVALENTS on page 200 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. The open circuit voltage Voc is Va -Vb , and the short circuit is the current ISC , from a to b. When a circuit is SHORT-CIRCUITED the current is said to be INFINITE.
STEPS FOR SOLVING THEVENIN EQUIVALENT CIRCUIT PROBLEMS: 1. Open the load resistor and remove the load resistance, R , from the circuit. L
2. Open current sources, and short voltage sources, by replacing the voltage source by its INTERNAL RESISTANCE.
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psst…don’t forget to take notesÎ
3. Calculate the equivalent resistance of the circuit looking from the load terminal towards the circuit. This is called the Thevenin’s Equivalent Resistance, R TH . 4. Calculate the SOURCE VOLTAGE, VS , by reconnecting the voltage sources, and finding the difference between the two voltage sources. 5. Calculate the SOURCE CURRENT, IS , using Ohm’s Law after finding the source voltage, VS , and the equivalent resistance with the source voltage connected, R S . 6. Calculate the OPEN CIRCUIT VOLTAGE, the Thevenin Voltage VTH , across each resistor, making sure to account for which voltage source is supplying the voltage to that resistor. This is done using Ohm’s Law. We can then calculate the voltage using the previously calculated source current in Step (5) and given resistance of the resistor. 7. Redraw the circuit with the calculated open circuit Voltage, VTH , in Step (6) as the voltage source and the calculated open circuit resistance R TH in step (3) as a series resistance. Connect the load resistor, R L , which we had removed in Step (1). This is the Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed by Thevenin’s Theorem. 8. We can now calculate the total current flowing through the circuit using Ohm’s Law expressed as: I T = 3|PREPINEER.COM
VTH . R TH + R L
psst…don’t forget to take notesÎ
CONCEPT EXAMPLE:
What is the Thevenin equivalent at terminal a-b for the circuit below? Calculate the total current flowing through the circuit.
SOLUTION:
1. Remove the load resistance, R L , from the circuit. In this problem, it is the 40Ω . 2. Open the current sources and short the voltages sources. The circuit would be drawn like the circuit below:
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3. Calculate the equivalent resistance looking form the load terminal towards the circuit. The 10Ω resistor is in parallel with the 20Ω resistor, so the equivalent resistance is calculated as:
R TH =
R 1 × R 2 20 × 10 = = 6.67Ω R 1 +R 2 20 + 10
4. Calculate the source voltage, VS , by reconnecting the two voltage sources in the circuit and taking the difference.
VS = V1 − V2 = 20V − 10V = 10V 5. Now, calculate the source current flowing in the circuit using Ohm’s Law. Plug inthe previously calculated source voltage and the equivalent resistance with the source voltage connected:
IS =
VS 20v-10v = = 0.33 amps R S 20!+10!
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6. Calculate the OPEN CIRCUIT VOLTAGE, the Thevenin Voltage, VTH , across each resistor using Ohm’s Law. Be sure to account for which voltage source is supplying the voltage to which resistor. Plugging in the previously calculated source current in Step (5) and given resistance of the resistor, calculate the voltage. The resistors are in series, so the current will be in the same in each of the resistors. Calculating the voltage drop across the 20Ω resistor with the 20 V voltage source:
(
)
VAB = V2 – R 20Ω x IS = VTH
(
)
VAB = 20 – 20" x 0.33 A = 13.33 volts Calculating the voltage drop across the 20Ω resistor with the 20 V voltage source:
(
)
VAB = V1 – R 10Ω x IS = VTH
(
)
VAB = 10 + 10" x 0.33amps = 13.33 volts
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psst…don’t forget to take notesÎ
The Thevenin Voltage is the same across each resistor:
VTH = 13.33Volts The Thevenin Equivalent Voltage is the same through each resistor. Just remember that since terminal a to terminal b is short circuited, the voltage cannot flow through it Therefore the calculation must use the voltage source that is directly connected to the resistor. 7. Redraw the circuit with the calculated open circuit Voltage, VTH , from Step (6) as the voltage Source. Then, connect the calculated open circuit resistance R TH from step (3) and the the load resistor,
R L , which we had removed in Step (1) in series. This is the
Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed by Thevenin’s Theorem.
8. We can now calculate the total current flowing through the circuit using Ohm’s Law expressed as: 7|PREPINEER.COM
psst…don’t forget to take notesÎ
IT =
VTH R TH + R L
This formula is not supplied in the Reference Handbook.
IT =
VTH 13.33v = = 0.286 amps R TH 6.67!+40!
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psst…don’t forget to take notesÎ
