Elec Class
learn electricity hands-on
Solved exercise · level 4 of 5
Skills: Analyse, Calculate
E1 (108 V) feeds through R1 = 2 Ω and E2 (24 V) through R2 = 8 Ω into a common junction; R3 = 6 Ω links that junction to the return. Write the loop and junction laws, then give the current in R1 (absolute value).
Parametric exercise: the values change with every draw; these are one example.
The three equations, ΣE − ΣRI = 0 around each loop plus the junction law:
(I) + 108 − 2·I₁ − 6·I₃ = 0
(II) − 24 + 8·I₂ + 6·I₃ = 0
(N) I₁ + I₂ − I₃ = 0
(N) gives I₁ = I₃ − I₂, call it (3), substituted into (I):
+ 108 − 2·(I₃ − I₂) − 6·I₃ = 0
⟺ + 108 − 2 I₃ + 2 I₂ − 6 I₃ = 0
⟺ + 108 + 2 I₂ − 8 I₃ = 0 (4)
Eliminate I₃ between (4) and (II):
(4) × 3: + 324 + 6 I₂ − 24 I₃ = 0
(II) × 4: − 96 + 32 I₂ + 24 I₃ = 0
+ 228 + 38 I₂ = 0, so I₂ = −228 / 38 = −6 A
Work back up:
(II): − 24 + 8·(−6) + 6·I₃ = 0, so I₃ = 12 A
(3): I₁ = I₃ − I₂ = 12 − (−6) = 18 A
Check at the junction: 18 − 6 = 12. Quick control: the junction potential is V_A = R₃·I₃ = 72 V. The answer boxes expect the absolute values: 18 A, 6 A and 12 A.