Thevenin Equivalent Voltage: why ignore the 3-kΩ resistor?
$begingroup$
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 Jan 4 at 16:55
Verbal Kint
3,3331412
asked Jan 4 at 16:23
fredfred
123110
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:55
Verbal Kint
3,3331412
asked Jan 4 at 16:23
fredfred
123110
$endgroup$
2
$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
1
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41
add a comment |
$begingroup$
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 Jan 4 at 16:55
Verbal Kint
3,3331412
asked Jan 4 at 16:23
fredfred
123110
$endgroup$
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 Jan 4 at 16:55
Verbal Kint
3,3331412
asked Jan 4 at 16:23
fredfred
123110
edited Jan 4 at 16:55
Verbal Kint
3,3331412
asked Jan 4 at 16:23
fredfred
123110
edited Jan 4 at 16:55
Verbal Kint
3,3331412
edited Jan 4 at 16:55
Verbal Kint
3,3331412
edited Jan 4 at 16:55
Verbal Kint
3,3331412
3,3331412
asked Jan 4 at 16:23
fredfred
123110
asked Jan 4 at 16:23
fredfred
123110
asked Jan 4 at 16:23
fredfred
123110
123110
2
$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
1
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41
add a comment |
2
$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
1
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41
2
2
$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
1
1
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.
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 Jan 4 at 16:58
P. ColladoP. Collado
1365
$endgroup$
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
add a comment |
$begingroup$
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 Jan 4 at 16:31
ShamtamShamtam
2,5331023
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
$endgroup$
add a comment |
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3 Answers
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$begingroup$
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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.
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 Jan 4 at 16:58
P. ColladoP. Collado
1365
$endgroup$
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
add a comment |
$begingroup$
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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.
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 Jan 4 at 16:58
P. ColladoP. Collado
1365
$endgroup$
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
add a comment |
$begingroup$
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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.
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 Jan 4 at 16:58
P. ColladoP. Collado
1365
$endgroup$
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 $Z_{TH} = 7kOmega + 3kOmega = 10kOmega$, 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 Kirchhoff'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 means 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 = 7k$Omega$.
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 Jan 4 at 16:58
P. ColladoP. Collado
1365
edited Jan 5 at 23:20
answered Jan 4 at 16:58
P. ColladoP. Collado
1365
answered Jan 4 at 16:58
P. ColladoP. Collado
1365
answered Jan 4 at 16:58
P. ColladoP. Collado
1365
1365
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
add a comment |
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Nice answer, but please please the "k" of the prefix kilo should be lower case and Kirchhoff has two h's ;-)
$endgroup$
– Massimo Ortolano
Jan 5 at 17:33
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
$begingroup$
Thanks a lot for the corrections. Is it ok now? ;)
$endgroup$
– P. Collado
Jan 5 at 23:21
add a comment |
$begingroup$
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 Jan 4 at 16:31
ShamtamShamtam
2,5331023
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:31
ShamtamShamtam
2,5331023
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:31
ShamtamShamtam
2,5331023
$endgroup$
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 Jan 4 at 16:31
ShamtamShamtam
2,5331023
answered Jan 4 at 16:31
ShamtamShamtam
2,5331023
answered Jan 4 at 16:31
ShamtamShamtam
2,5331023
answered Jan 4 at 16:31
ShamtamShamtam
2,5331023
2,5331023
add a comment |
add a comment |
$begingroup$
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 Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
$endgroup$
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 Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
edited Jan 4 at 17:21
answered Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
answered Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
answered Jan 4 at 16:32
Spehro PefhanySpehro Pefhany
210k5161425
210k5161425
add a comment |
add a comment |
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$begingroup$
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
$endgroup$
– thece
Jan 4 at 16:36
1
$begingroup$
The Thev, Equiv cct. uses 3k + 7k=10k in series with 8m*7k=56V
$endgroup$
– Sunnyskyguy EE75
Jan 4 at 16:41