- #1
Marcin H
- 306
- 6
Homework Statement
Find the voltage, V, across the 6A current source
Homework Equations
V=IR
Node Voltage Method
The Attempt at a Solution
Did I set this up correctly to find my voltage?
Oh, woops. Read the polarities incorrectly. So if I change the last 2 to positive, it's good?gneill said:Almost. Be careful of the signs of the terms.
Judging by the sign you gave to the 6A supply term you're summing currents leaving the node. So why are your last two terms negative?
Thanks!gneill said:Yes, that would be a correct expression. Make it an equation by setting it equal to zero
Got stuck again! The next question asks "What is the power associated with the 10V source?" I can find it using P=IV=I^2R=V^2/r, but how do I find the current through the 10V source?gneill said:Yes, that would be a correct expression. Make it an equation by setting it equal to zero
Ohh. So I use (Vx-10)/20= (36-10)/10 = 2.6A And then using P=IV I can do (2.6A)(10V)= 26W***?gneill said:Each term in the node equation represents a branch current. One of the terms corresponds to that current. Pick it out and use it now that you've found the node voltage.
Ok thanks! 13W***gneill said:That would be the idea, yes.
Marcin H said:Ohh. So I use (Vx-10)/20= (36-10)/10 = 1.3A And then using P=IV I can do (1.3A)(10V)= 13W***?
Woops! 26 watts.gneill said:Check your calculation for the current. (36 - 10)/10 = ?
Try again. Show your calculations.Marcin H said:Woops! 26 watts.
What? Where did I go wrong? I = (36-10)/10 = 26/10 = 2.6A. Using that current and P=IV I can find the power in the 10V source. P = (2.6A)(10V) = 26 Wattsgneill said:Try again. Show your calculations.
Marcin H said:What? Where did I go wrong? I = (36-10)/10 = 26/10 = 2.6A. Using that current and P=IV I can find the power in the 10V source. P = (2.6A)(10V) = 26 Watts
Ah. My mistake. Sorry. I was thinking of the power associated with the branch as a whole. You have the correct result.Marcin H said:ORIGINAL QUESTION: what is the power associated with the 10V source?
The Node Voltage Method is a circuit analysis technique used to find the voltage at each node in a circuit. It involves setting up a system of equations based on Kirchhoff's Current Law (KCL) and solving for the unknown node voltages.
To apply the Node Voltage Method to a circuit with a 6A current source, you would first label all the nodes in the circuit. Then, you would use KCL to set up equations at each node with unknown voltages. The equation for the node connected to the 6A current source would include a term for the 6A current source. Finally, you would solve the system of equations to find the voltages at each node, including the voltage across the 6A current source.
Yes, the Node Voltage Method can be used for circuits with multiple current sources. Each current source would be represented as a term in the KCL equation at its corresponding node. The system of equations can then be solved to find the voltages at each node, including the voltage across each current source.
The Node Voltage Method is advantageous because it is a systematic and methodical approach to analyzing circuits, making it less prone to errors. It also allows for easy application of KCL, which is a fundamental law in circuit analysis. Additionally, the Node Voltage Method can be used for circuits with both independent and dependent sources.
One limitation of the Node Voltage Method is that it can only be used for circuits with a single reference node. Additionally, it can become more complex and time-consuming to apply for circuits with a large number of nodes. In these cases, other circuit analysis techniques such as the Mesh Current Method may be more efficient.