1. Gustav Robert Kirchhoff
Kirchhoff's Laws
Calculate the currents
1. I1;
2. I2; and
3. I3.
E1 = 18 V
8 Ω
5 Ω
E2 = 12 V I2 I1 I3
1.5 Ω a b c d e f
Gustav Robert Kirchhoff

2. Gustav Robert Kirchhoff
Kirchhoff's Laws
Calculate the currents
1. I1;
2. I2; and
3. I3.
E1 = 18 V
8 Ω
5 Ω
E2 = 12 V I2 I1 I3
1.5 Ω a b c d e f
Gustav Robert Kirchhoff

3. Apply Kirchhoff's voltage law to the loop
Kirchhoff's Laws
E1 = 18 V b c d e I3 I2 I1
5 Ω
1.5 Ω
Apply Kirchhoff's voltage law to the loop
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∑(potential differences) = 0
18 V - V5 Ω - 12 V - V8 Ω = 0
∑(potential rises) = ∑(potential drops)
V = V5 Ω + V8 ΩΩ
8 Ω
6 V = 5 I2 + 8 I1
E2 = 12 V
Apply Kirchhoff's voltage law to the loop
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Apply Kirchhoff's current law at b
∑(current in) = ∑(currents out)
∑(potential rises) = ∑(potential drops)
-12 V = V1.5Ω + V8 Ω
I1 = I2 + I3
-12 V = 1.5 I3 + 8 I1

4. Kirchhoff's Laws
From
previous slide
E1 = 18 V
E2 = 12 V I2 I1 I3 a b c d e f
6 V = 8 I1 + 5 I2
-12 V = 8 I I3
I1 = I2 + I3
substitute
I1 = I2 + I3
into
-12 V = 8 I I3
8 Ω
1.5 Ω
5 Ω
-12 V = 8 I (I1 - I2)
-12 V = 8 I I I2
-12 V = 9.5 I I2
Using simultaneous equations
-12 V = 9.5 I I2
6 V = 8 I1 + 5 I2
-60 V = 47.5 I I2
9 V = 12 I I2
-51 V = 59.5 I1
I1 = A

5. Excellence E1 = 18 V E2 = 12 V a b c d e f 8 Ω 1.5 Ω 5 Ω
Kirchhoff's Laws
E1 = 18 V
E2 = 12 V I2 I1 I3 a b c d e f
From
previous slide
6 V = 8 I1 + 5 I2
-12 V = 8 I I3
-12 V = 9.5 I I2
I1 = A
8 Ω
1.5 Ω
5 Ω
substitute
I1 = A
into
-12 V = 9.5 I I2
-12 V = (9.5 x -0.86) I2
= I2
I2 = 2.5 A
Excellence
substitute calculated
currents into
I3 = I1 – I2
I3 = – 2.5
I3 = -3.4 A
1. I1 = A
2. I2 = 2.5 A
3. I3 = -3.4 A