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Circuit 1: I would use superposition. The theory is explained here. Make three equivalent circuits which each contain only one source (10V, 8V, 2A). solve the circuits and add up the results.
Circuit 2: Since a-b is an open circuit, no current flows and the 2Ohm resistor at the output is without effect. You have here a dependant voltage source (2vx). Therefore simple superposition will not work. Set up the current and voltage equations for the network and solve the resulting system of linear equations.
Circuit 2: Since a-b is an open circuit, no current flows and the 2Ohm resistor at the output is without effect. You have here a dependant voltage source (2vx). Therefore simple superposition will not work. Set up the current and voltage equations for the network and solve the resulting system of linear equations.
Hi Harald Kapp,
Million thanks for your reply, for
circuit 1 to how find equivalent circuits at 2A ?
Circuit 2 I not very understand can brief more deep ?
Million Thanks
For teh 2A case, the voltage sources are replaced by short circuits. You can then evaluate the remaining resistor combinations (as series or parallel circuits) and replace each combination by a single resistor with an equivalent value.
For example: the 2*6Ohm resistors to the right can be replaced by a single 3Ohm resistor.
Circuit 2: Use Kirchoff's laws:
- for each node in the circuit the sum of all currents into and out of that node is 0A.
- for each closed loop in the circuit the sum of all voltages along the loop is 0V.
It is important that you label all currents and voltages in a unique and easily identifiable manner. You could, for example, label the voltage across the right 6Ohm resistor label V6 and the current accordingly I6. The name is not important, only that you can identify all variables. With that name scheme, you'd get V6=I6*6Ohm as one equation. You need to set up enough equations to solve the system of linear equations resulting from this effort.
I won't do all the math for you, this hint should lead you on the right track.
For example: the 2*6Ohm resistors to the right can be replaced by a single 3Ohm resistor.
Circuit 2: Use Kirchoff's laws:
- for each node in the circuit the sum of all currents into and out of that node is 0A.
- for each closed loop in the circuit the sum of all voltages along the loop is 0V.
It is important that you label all currents and voltages in a unique and easily identifiable manner. You could, for example, label the voltage across the right 6Ohm resistor label V6 and the current accordingly I6. The name is not important, only that you can identify all variables. With that name scheme, you'd get V6=I6*6Ohm as one equation. You need to set up enough equations to solve the system of linear equations resulting from this effort.
I won't do all the math for you, this hint should lead you on the right track.
Hi Harald Kapp,
Million thanks again for your reply and guide, I solve the circuit 1 follow the guide and hits that you provide but next step how to solve the circuit using Thevenin Theorem ? Here are my answer for using superposition Theorem:-
VS1~10V = VR1 + VR3 = 4.5 + 5.5V =10 V
VS2~8V= - VR2+VR3 = 5V
VS3.1~2V = VR1+VR3=2V
VS3.2~2V = VR2+VR3=1V
Total VS2~8V = 5+2+1= 8V
To verify your equations you should show us a picture of the equivalent circuits you used and where the voltages are labeled. But you want to use the Thevenin theorem, so I will not follow that venue further.
You don't describe this in your OP but I assume that only the current through "R" or the respective voltage across it is of interest.
To apply the Thevenin theorem (without superposition), you have a few different options.
For example you can decide to simplify the circuit by using only voltage sources.
- Take the 8V source to the right plus the associated 2*6Ohm resistors. You can replace it by a single source+1 resistor.
- Take the 2A source with its 1Ohm resistor. Replace it by a voltage source with equivalent series resistor, too.
(No image for these two simple steps).
Now take these two new equivalent sources and combine them to another single voltage source with series resistor. Your circuit looks like this:
Note that I removed all values on purpose - it's your job to calculate them.
You can now go one step further and combine the two voltage sources to a single one like this:
This is the point where you are ready to calculate the voltage across R or the current through R. Or both, of course.
You can use the same technique for the second circuit.
You don't describe this in your OP but I assume that only the current through "R" or the respective voltage across it is of interest.
To apply the Thevenin theorem (without superposition), you have a few different options.
For example you can decide to simplify the circuit by using only voltage sources.
- Take the 8V source to the right plus the associated 2*6Ohm resistors. You can replace it by a single source+1 resistor.
- Take the 2A source with its 1Ohm resistor. Replace it by a voltage source with equivalent series resistor, too.
(No image for these two simple steps).
Now take these two new equivalent sources and combine them to another single voltage source with series resistor. Your circuit looks like this:
Note that I removed all values on purpose - it's your job to calculate them.
You can now go one step further and combine the two voltage sources to a single one like this:
This is the point where you are ready to calculate the voltage across R or the current through R. Or both, of course.
You can use the same technique for the second circuit.
Attachments
To verify your equations you should show us a picture of the equivalent circuits you used and where the voltages are labeled. But you want to use the Thevenin theorem, so I will not follow that venue further.
You don't describe this in your OP but I assume that only the current through "R" or the respective voltage across it is of interest.
To apply the Thevenin theorem (without superposition), you have a few different options.
For example you can decide to simplify the circuit by using only voltage sources.
- Take the 8V source to the right plus the associated 2*6Ohm resistors. You can replace it by a single source+1 resistor.
- Take the 2A source with its 1Ohm resistor. Replace it by a voltage source with equivalent series resistor, too.
(No image for these two simple steps).
Now take these two new equivalent sources and combine them to another single voltage source with series resistor. Your circuit looks like this:
Note that I removed all values on purpose - it's your job to calculate them.
You can now go one step further and combine the two voltage sources to a single one like this:
This is the point where you are ready to calculate the voltage across R or the current through R. Or both, of course.
You can use the same technique for the second circuit.
Hi Harald Kapp,
Thanks for the reply, And apologies if cause trouble for you. I follow the instruction that given, is the V2 that I add is correct value?
