Kirchhoff’s Current Law (KCL)
KCL #1
ƩI = 0
I 1 + I2 + I3 – I 4 – I 5 – I 6 = 0
KCL #2
ƩIin = ƩI out
I 1 + I2 + I3 = I4 + I5 + I 6
Example 1
Given:
I1 = 15 A I4 = 8 A
I2 = 10 A I5 = 10 A
I3 = 7 A
Req’d: Verify KCL #1 & #2
Sol’n:
KCL #1
ƩI = 0
I 1 + I2 – I 3 – I 4 – I 5 = 0
15 + 10 – 7 – 8 – 10 = 0
25 – 25 = 0
0=0
KCL #2
ƩIin = ƩI out
I 1 + I2 = I 3 + I4 + I5
15 + 10 = 7 + 8 + 10
25 A = 25 A
Given:I1 = 10 A
Req’d: I4
Sol’n:
I2 = 20 A
I3 = 5A
KCL #2
ƩIin = ƩI out
I 1 + I3 + I 4 = I 2
10 + 5 + I4 = 20
I4 = 20 – 15
I4 = 5 A
KCL #1
ƩI = 0
I 1 + I3 + I 4 - I 2 = 0
10 + 5 + I4 – 20 = 0
I4 – 5 = 0
I4 = 5 A
At node C
KCL #1
I1 + 4 – I2 – 3 = 0
I1 – I2 + 1 = 0
I1 = I2 – 1
At node B
KCL #1
I2 + 3 – 4 – 2 – I3 = 0
I2 – I3 – 3 = 0
I2 = I3 + 3
At node A
KCL #1
I3 + 1 + 4 – 5 = 0
I3 + 5 – 5 = 0
I3 = 0
I3+4 – 1 – 5 = 0
I3 + 4 – 6 = 0
I3 = 2 A
Substitute I3 = 0 in
I2 = I3 + 3
I2 = 0 + 3
I2 = 3 A
Substitute I2 = 3 A in
I1 = I2 -1
I1 = 3 – 1
I1 = 2 A
Substitute I3 = 0 in
I2 = I3 + 3
I2 = 2 + 3
I2 = 5 A
Substitute I2 = 5 A in
I1 = I2 -1
I1 = 5 – 1
I1 = 4 A