Both are fundamental "laws" that always hold. In terms of approaching a given problem sometimes one is more straightforward or advantageous in terms of the mathematical complexity. For example, there might be three loops but only two essential nodes in a circuit, so using KVL loops would require three equations to find the loop current, but only two node equations to find the node voltages. However, neither method would work at all if both KVL and KCL didn't hold for the circuit.goonking said:
Circuit question -- Two sources and two resistors....
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
36 replies · 7K views
monski24
- 1
- 0
what is the answer for this problem. can you give us the steps how to deal with this
Gold Member
- 5,661
- 4,508
Just a clarification here.goonking said:
The Controlled Current Source (CCS) V0 is sensing the voltage across the 4 Ohm resistor. This is shown by the V0 symbol below the resistor. This is a differential measurement, i.e. fully floating with respect to any Ground, and conceptually uses two wires.
The wires are typically not drawn in the schematic to reduce clutter. The "+" and "-" symbols at the resistor show the polarity connection of the two sense wires for the CCS.
Hope this helps!
Tom
boo
- 26
- 8
Forget Vo for a minute. Just apply KCL at that top node (junction of the 2 current sources and the 4 ohm resistor). Take the bottom as the reference node. We then have (V - 0)/(4 + 6) -10 - 2*Vo =0. Now Vo = 4*(0 - V)/(4 + 6). Substitute for Vo and solve for V. Finally, since the current is entering the negative terminal of the dependent current source the power absorbed will be equal to - (V)*(2*Vo). (Hint Vo will actually be negative in this case.)
boo
- 26
- 8
Let me also point out that the main reason you are getting stuck on this problem is because you are taking Vo as the voltage at that top node which it is not. The left going current is neither Vo/4 nor Vo/(4 + 6). The correct (left going) current is simply V/(4 + 6) where V, as above, is simply the voltage at that essential node on top.
Jody
Gold Member
- 454
- 230
This thread is super old. Someone has already answered the question and another person bumped it way later probably not reading the old responses.boo said:
I wouldn't have followed up, but I did want to add something: I understand that you are correct about the "left going" current not being Vo/4, but the current in that left branch IS Vo/4 and can still be easily used for KCL. If it's really important you can say the "left going" current is -Vo/4. Sure you don't know what the voltage is at the top, but you can use Ohm's law on that resistor that has Vo assigned across it. This works because the two resistors are in series, and so you know the same current is flowing through it. So long as current is flowing through that branch the 6 Ohm resistor does not affect Vo.
The eye opener to that problem is that it would affect the power (second part of the problem). Finding the voltage at the top node would then become helpful because the second part of the question asks to find the power. I think considering Vo/4 from the start is easier and you can use that current to easily get the voltages without considering a voltage divider or careful bookkeeping of the polarities.
boo
- 26
- 8
Whoa! Didn't realize that this was an ancient thread. Would not waded in. But agree that left going current is - Vo/4.