1. Kirchhoff – Step by step
Step 1 Look at the information given Spot the obvious! I1 I2
The voltage across this resistor ΔV = IR = I1 x 30 = 30 I1

2. Kirchhoff – Step by step
Step 1 Look at the information given Spot the obvious! I1 I2 I3
The currents at this junction I1 + I2 + I3 = 0
either I1 or I2 may be negative as one is going out of the junction

3. Kirchhoff – Step by step
Step 2 break the circuit into a set of Closed Loops & label them! I1 I2 A E D B F C

4. Kirchhoff – Step by step
Step 3 Decide which way to go around the loop. . . I1 I2 A E D B F C

5. Kirchhoff – Step by step
Step 4 Apply the rules LOOP ABCD
ε = +18 V
ε = +6 V I1 I2 A E D
This resistor is not in the loop –ignore it !
ΔV = -I1R = -30I1 B F C
Do the sum! – 30 I1 = 0
30 I1 = → I1 = 0.8 A

6. Kirchhoff – Step by step
Step 5 Apply the rules LOOP ABFE
ε = +18 V I1 I2 A E D
I1 = 0.8 A
The voltage across this resistor is I3 x R
ΔV = -30 I3 I3
ΔV = -30 I1 = -24 V B F C
Do the sum! I3 = N.B. Negative current
30 I3 = → I3 = -6 ÷ 30 = A flowing E > F

7. Kirchhoff – Step by step
Step 6 Apply the rules Current at junction E I1 I2 A E D
I1 = 0.8 A I3
Currents entering a junction are positive + I
Currents leaving a junction negative - I
I3 = 0.2 A B F C
Do the sum I1 + I2 + I3 = 0
- 0.8 – I3 = 0 → I3 = = 1.0 A