Engineering How Do You Solve Basic Electric Circuit Analysis Problems with KCL and KVL?

[wompkins]

Homework Statement

From this circuit diagram assume Vs and R1 through R2 are known.
Find the following:
a) How many unknown currents are there?
b) How many independent equations can be written using KCL?
c) How many independent equations can be derived from KVL equations?

Homework Equations

KCL: Current in - Current out = 0
KVL: Voltage drop around loop should be zero
Ohm's Law: V = iR

The Attempt at a Solution

This problem should not be that difficult but the part that is tripping me up is the middle wire with no resistance in it. Am I correct in assuming that no current will be flowing through it? If so these are the answers I got.

a) 3 unknown currents
b) 1 KCL equation
c) 2 KVL equation

I got (a) from just drawing currents on diagram.

I got (b) from node-voltage which says:
# of KCL eqns = (essential nodes - 1)

Lastly, (c) from mesh-current which says:
# of KVL eqns = branches - (nodes - 1)

 

[phyzguy]

Why do you think the currents through R2 and R3 are equal?

 

[AlephZero]

Am I correct in assuming that no current will be flowing through it?

You can draw the circuit diagram without that piece of wire. Just join the ends of R2 R3 R6 and R7 together.

If you solve that circuit to find the currents in those four resistors, count the number of essential nodes, etc, you can answer your own question.

 

[gneill]

You can draw the circuit diagram without that piece of wire. Just join the ends of R2 R3 R6 and R7 together.

Unless, of course, that wire happens to represent an actual wire in a physical actualization of the circuit. In which case you could think of it as a resistor with value ~0 Ohms for purposes of analyzing the circuit.

 

[wompkins]

You can draw the circuit diagram without that piece of wire. Just join the ends of R2 R3 R6 and R7 together.

If you solve that circuit to find the currents in those four resistors, count the number of essential nodes, etc, you can answer your own question.

Thank you! I can solve it from there