Thevenin Equivalent Voltage: why ignore the 3-kΩ resistor?
In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)
passive-networks thevenin
edited yesterday
Verbal Kint
3,2291312
asked yesterday
fred
11219
add a comment |
In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)
passive-networks thevenin
edited yesterday
Verbal Kint
3,2291312
asked yesterday
fred
11219
2
The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
1
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday
add a comment |
In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)
passive-networks thevenin
edited yesterday
Verbal Kint
3,2291312
asked yesterday
fred
11219
In order to find the Thevenin voltage here, why is it that the 3k Οhm resistor is ignored and no current goes through it? (hence $V_{TH}=8mA cdot 7kΩ = 56V$)
passive-networks thevenin
passive-networks thevenin
edited yesterday
Verbal Kint
3,2291312
asked yesterday
fred
11219
edited yesterday
Verbal Kint
3,2291312
asked yesterday
fred
11219
edited yesterday
Verbal Kint
3,2291312
edited yesterday
Verbal Kint
3,2291312
edited yesterday
Verbal Kint
3,2291312
3,2291312
asked yesterday
fred
11219
asked yesterday
fred
11219
asked yesterday
fred
11219
11219
2
The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
1
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday
add a comment |
2
The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
1
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday
2
2
The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
1
1
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday
add a comment |
3 Answers
3
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Computing a Thevenin Equivalent requires two steps:
- Obtain the Thevenin Impedance $Z_{TH}$
- Obtain the Thevenin Voltage $V_{TH}$
In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.
In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!
Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
$$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
$$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7KOhms.
This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.
simulate this circuit – Schematic created using CircuitLab
answered yesterday
P. Collado
1063
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The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.
answered yesterday
Shamtam
2,4431023
add a comment |
The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).
When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.
answered yesterday
Spehro Pefhany
203k4150408
add a comment |
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3 Answers
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Computing a Thevenin Equivalent requires two steps:
- Obtain the Thevenin Impedance $Z_{TH}$
- Obtain the Thevenin Voltage $V_{TH}$
In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.
In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!
Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
$$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
$$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7KOhms.
This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.
simulate this circuit – Schematic created using CircuitLab
answered yesterday
P. Collado
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Computing a Thevenin Equivalent requires two steps:
- Obtain the Thevenin Impedance $Z_{TH}$
- Obtain the Thevenin Voltage $V_{TH}$
In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.
In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!
Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
$$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
$$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7KOhms.
This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.
simulate this circuit – Schematic created using CircuitLab
answered yesterday
P. Collado
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Computing a Thevenin Equivalent requires two steps:
- Obtain the Thevenin Impedance $Z_{TH}$
- Obtain the Thevenin Voltage $V_{TH}$
In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.
In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!
Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
$$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
$$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7KOhms.
This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.
simulate this circuit – Schematic created using CircuitLab
answered yesterday
P. Collado
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Computing a Thevenin Equivalent requires two steps:
- Obtain the Thevenin Impedance $Z_{TH}$
- Obtain the Thevenin Voltage $V_{TH}$
In order to compute $Z_{TH}$ you have to "turn off" the independent current source. As it is a current generator you will need to place an open circuit there. This will "disconnect" the 4k resistor from the circuit, giving you $Z_{TH} = 7K + 3K = 10K$, as the resistors are in series.
In order to obtain $V_{TH}$ you need an open circuit between the terminals A and B. Having this open circuit will cause no current to flow through the 3K resistor, as it would violate Kirchoff's First Law. That current would have no way to get "back" to the circuit, so you would be diminishing the electron count!
Another way to look at it is as if you had an infinite resistor between A and B. The air acts as an insulator, so given a real valued voltage drop in A and V $V_{AB} = alpha$ you would get:
$$I = lim_{R to infty}frac{V_{AB}}{R} = lim_{R to infty}frac{alpha}{R} = 0$$
No matter how you want to look at it, you have no current flowing there. This leaves the current flowing only in the left loop. Having no current in the 3K resistor measn we will have no voltage drop there. Remember that we want to find $V_{TH} = V_A - V_B$. Following the circuit we see that the only drop involved is the one in the 7K resistor. By Ohm's Law, that drop will be $V_{DROP} = I·R$, so:
$$V_A = V_B - I·R to V_A - V_B = V_{TH} = - I·R$$
Being I = 8mA and R = 7KOhms.
This is the Thevenin Equivalent and why the 3K resistor doesn't play a role in the Thevenin Voltage.
simulate this circuit – Schematic created using CircuitLab
answered yesterday
P. Collado
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
P. Collado
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
P. Collado
1063
answered yesterday
P. Collado
1063
1063
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
P. Collado is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.
answered yesterday
Shamtam
2,4431023
add a comment |
The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.
answered yesterday
Shamtam
2,4431023
add a comment |
The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.
answered yesterday
Shamtam
2,4431023
The Thévenin voltage is also known as the "open-circuit" voltage. In this case, the open circuit is between points A and B. Because it's an open circuit, no current flows through the 3k resistor, and thus there is no voltage drop across it.
answered yesterday
Shamtam
2,4431023
answered yesterday
Shamtam
2,4431023
answered yesterday
Shamtam
2,4431023
answered yesterday
Shamtam
2,4431023
2,4431023
add a comment |
add a comment |
The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).
When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.
answered yesterday
Spehro Pefhany
203k4150408
add a comment |
The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).
When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.
answered yesterday
Spehro Pefhany
203k4150408
add a comment |
The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).
When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.
answered yesterday
Spehro Pefhany
203k4150408
The Thevenin voltage is when the output terminals are open-circuit so any series resistance (3K in this case) has no effect. We can also ignore the 4K resistor since it's in series with a current source, and the open-circuit voltage is $8mA cdot 7K$ = 56V (A is negative wrt B).
When you go to calculate the equivalent series resistance from that voltage, the 3K will come into play. You can apply the current divider formula to find the current through the 3K (5.6mA) and thus the output resistance. Or just replace voltage sources with shorts and current sources with open circuits and look at the resistance looking in from the output.
answered yesterday
Spehro Pefhany
203k4150408
edited 23 hours ago
answered yesterday
Spehro Pefhany
203k4150408
answered yesterday
Spehro Pefhany
203k4150408
answered yesterday
Spehro Pefhany
203k4150408
203k4150408
add a comment |
add a comment |
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The equivalent voltage Vth is the voltage obtained at terminals A-B of the network with terminals A-B open circuited. en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
– thece
yesterday
1
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
– Tony EE rocketscientist
yesterday