Kirchhoff’s two laws
Answers are for parents and teachers.
1
Wires meet at a junction. 6 A and 1.5 A flow in, 2 A and 3.5 A flow out, and a fifth wire is assumed to carry I out of the junction. Find I and the real direction.
6 + 1.5 = 2 + 3.5 + I, so I = 2 A. It is positive, so 2 A really flows out of the junction.
2
Use the layout of Lesson 3 with E₁ = 17 V, R₁ = 2 Ω, E₂ = 21 V, R₂ = 3 Ω and R₃ = 5 Ω (no internal resistances). Find I₁, I₂, I₃ and the voltage across R₃.
Loop A: 17 = 2I₁ + 5(I₁ + I₂) = 7I₁ + 5I₂. Loop B: 21 = 3I₂ + 5(I₁ + I₂) = 5I₁ + 8I₂. Multiply A by 8 and B by 5: 136 = 56I₁ + 40I₂ and 105 = 25I₁ + 40I₂. Subtract: 31 = 31I₁, so I₁ = 1 A. Then 17 = 7 + 5I₂ gives I₂ = 2 A, and I₃ = 3 A. Across R₃: 3 × 5 = 15 V.
3
Check the answers of the previous question with the outer loop E₁ − E₂ = I₁R₁ − I₂R₂.
17 − 21 = −4. And 1 × 2 − 2 × 3 = 2 − 6 = −4 ✓.
4
A solution gives −0.8 A for a branch with a 5 Ω resistor, where the assumed current ran from left to right. Give the real direction and the voltage across the resistor.
The real current is 0.8 A from right to left. The voltage across the resistor is 0.8 × 5 = 4 V.
5
Name the conservation law behind each of Kirchhoff’s two laws and say in one sentence how each is used when solving a circuit.
First law: conservation of charge; it gives one equation at a junction (currents in = currents out). Second law: conservation of energy; it gives one equation for each independent loop (sum of emfs = sum of IR).