Egypt Secondary 3 Physics, Week 3: Solving a two-loop circuit step by step

Kirchhoff’s two laws

Week 3 · Chapter 1 (C): Kirchhoff’s first law (conservation of charge, the node rule) and second law (conservation of energy, the loop rule), the guidelines for assumed current directions and signs, and three worked multi-loop circuits · 3 lessons

Solving a two-loop circuit step by step

Physics Sec 3 Term 1, Chapter 1: pp. 18–23

GoalSolve a circuit with two loops for its three currents with one node equation and two loop equations.

New words

shared branch · the branch that belongs to both loopssimultaneous equations · two or more equations with the same unknowns, solved togetherelimination · adding or subtracting equations so that one unknown disappears

Explanation

A circuit with batteries in different branches cannot be cut into series and parallel parts, but Kirchhoff’s laws solve it. Take the layout in the table: two junctions P (top) and Q (bottom) are joined by three branches, and each battery has its positive pole on the P side. Loop A goes up branch 1 and down branch 3; loop B goes up branch 2 and down branch 3. Steps: (1) assume a current in every branch; (2) write the node equation at P; (3) write one loop equation for loop A and one for loop B with the sign rules; (4) put the node equation into them to leave two equations with two unknowns; (5) solve by elimination; (6) check the answers in an equation you did not use. A negative current means that the real direction is opposite to the assumed one.

Branch	Contains	Assumed current
1 (left)	battery E₁ (+ pole up) and R₁	I₁ upward, from Q to P
3 (middle)	resistor R₃ only	I₃ downward, from P to Q
2 (right)	battery E₂ (+ pole up) and R₂	I₂ upward, from Q to P

Worked examples

E₁ = 12 V with R₁ = 2 Ω, E₂ = 6 V with R₂ = 3 Ω, and R₃ = 6 Ω, all emfs without internal resistance. Node: I₃ = I₁ + I₂. Loop A: 12 = 2I₁ + 6I₃ = 8I₁ + 6I₂. Loop B: 6 = 3I₂ + 6I₃ = 6I₁ + 9I₂. Divide A by 2 and B by 3: 6 = 4I₁ + 3I₂ and 2 = 2I₁ + 3I₂. Subtract: 4 = 2I₁, so I₁ = 2 A. Then 3I₂ = 2 − 4, so I₂ = −2/3 A (about −0.67 A), and I₃ = 2 − 2/3 = 4/3 A (about 1.33 A). Check loop A: 2 × 2 + 6 × 4/3 = 4 + 8 = 12 ✓. Check loop B: 3 × (−2/3) + 8 = 6 ✓. I₂ is negative, so the real current in branch 2 flows downward, into the positive pole of battery 2: this battery is being charged. The voltage across R₃ is 6 × 4/3 = 8 V.

E₁ = 14 V with R₁ = 1 Ω, E₂ = 15 V with R₂ = 3 Ω, and R₃ = 4 Ω. Loop A: 14 = I₁ + 4(I₁ + I₂) = 5I₁ + 4I₂. Loop B: 15 = 3I₂ + 4(I₁ + I₂) = 4I₁ + 7I₂. Multiply A by 7 and B by 4: 98 = 35I₁ + 28I₂ and 60 = 16I₁ + 28I₂. Subtract: 38 = 19I₁, so I₁ = 2 A. Then 14 = 10 + 4I₂ gives I₂ = 1 A, and I₃ = 3 A. Check: 14 = 2 + 12 ✓ and 15 = 3 + 12 ✓.

A free check for Example 1 is the outer loop: up branch 1 and down branch 2. It crosses battery 1 from − to + (+E₁) and battery 2 from + to − (−E₂). In branch 1 the assumed current agrees with the loop (+I₁R₁), in branch 2 it runs against it (−I₂R₂). So E₁ − E₂ = I₁R₁ − I₂R₂: 12 − 6 = 6 and 2 × 2 − 3 × (−2/3) = 4 + 2 = 6 ✓.

Class activity

Design a circuit that you can solve. Choose the currents I₁ = 1 A, I₂ = 2 A and I₃ = 3 A, and the resistors R₁ = 1 Ω, R₂ = 2 Ω and R₃ = 3 Ω. Then find the emfs that give them with E₁ = I₁R₁ + I₃R₃ and E₂ = I₂R₂ + I₃R₃. Give the circuit, without the answers, to a friend and let them solve it. Paper only.

Practice

Use the layout of the lesson with E₁ = 9 V, R₁ = 1 Ω, E₂ = 4 V, R₂ = 2 Ω and R₃ = 3 Ω (no internal resistances). Find I₁, I₂ and I₃.

Check the answer of the previous question with the outer loop equation E₁ − E₂ = I₁R₁ − I₂R₂.

In the same circuit, what does the sign of I₂ tell you?

In the same circuit, find the voltage across R₃ and say which of P and Q is at the higher potential.