- it is a technique which relies on KVL.
Consider this is a circuit, with two meshes 1 and 2
- Next, assume current flow clockwise(CW) for easier.
- Define two mesh current I1 and I2.
- Then, assign voltage polarities to all resistor.
- Use KVL in loop 1 and 2 to get the equation.
- But, in R3 there are two current flow I1 and I2
- So, we write it in this manner -R3(I1-I2) for loop one and for loop two - R3(I2-I1)
- Loop 1 equation: 3 - R1I1 - R3(I1-I2) = 0
- For Loop 2 equation: 10 - R3(I2-I1) - R2I2 = 0
- We can now substitute the value of the resistors, and expand it.
- Use MATRIX to find the value of I1, I2
- For I3, you may use KCL from this node:
I1 = I3 + I2
I3 = I1 - I2
I3 = 2 - 3
I3 = -1 mA
No comments:
Post a Comment