Topics Covered in this ppt
Kirchhoff’s Current Law (KCL)
Kirchhoff’s Voltage Law (KVL)
Method of Branch Currents
Method of Mesh Currents
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
The 6-A IT into point C divides into the 2-A I3 and 4-A I4-5
I4-5 is the current through R4 and R5
IT − I3 − I4-5 = 0
6A − 2A − 4A = 0
At either point C or point D, the sum of the 2-A and the 4-A branch currents must equal the 6A line current.
Therefore, Iin = Iout
In Figure 9-2, for the inside loop with the source VT, going counterclockwise from point B,
90V + 120V + 30V = 240V
If 240V were on the left side of the equation, this term would have a negative sign.
The loop equations show that KVL is a practical statement that the sum of the voltage drops must equal the applied voltage.
For the loop CEFDC without source the equation is
Application of Kirchhoff’s laws to a circuit with two sources in different branches.
VR1 = I1R1
VR2 = I2R2
VR3 = I3R3
VR3 = (I1+I2)R3
V1 – I1R1 – (I1+I2) R3 = 0
V2 – I2R2 – (I1+I2) R3 = 0
Solving for currents
Using the method of elimination, multiply the top equation by 3 to make the I2 terms the same in both equations
9I1 + 3I2 = 42
1I1 + 3I2 = 7
7I1 = 35
I1 = 5A
To determine I2, substitute 5 for I1
2(5) + 3I2 = 7
3I2 = 7 − 10
3I2 = −3
I2 = −1A
This solution of −1A for I2 shows that the current through R2 produced by V1 is more than the current produced by V2.
The net result is 1A through R2 from C to E
Calculating the Voltages
VR1 = I1R1 = 5 x 12 = 60V
VR2 = I2R2 = 1 x 3 = 3V
VR3 = I3R3 = 4 x 6 = 24V
Note: VR3 and VR2 have opposing polarities in loop 2.
This results in the −21V of V2
Checking the Solution
At point C: 5A = 4A + 1A
At point D: 4A + 1A = 5A
Around the loop with V1 clockwise from B,
84V − 60V − 24V = 0
Around the loop with V2 counterclockwise from F,
21V + 3V − 24V = 0
Fig. 9-7: Method of node-voltage analysis for the same circuit as in Fig. 9-5.
Use either the rules for meshes with mesh currents or the rules for loops with branch currents, but do not mix the two methods.
To eliminate IB and solve for IA, divide the first equation by 2 and the second by 3. then
9IA − 3IB = 42
−2IA + 3IB = −7
Add the equations, term by term, to eliminate IB. Then
7IA = 35
IA = 5A
To calculate IB, substitute 5 for IA in the second equation:
−2(5) + 3IB = −7
3IB = −7 + 10 =3
IB = 1A
The positive solutions mean that the electron flow for both IA and IB is actually clockwise, as assumed.
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