Kirchhoff’s Rules & Multiloop Circuits
- Reduce networks with series and parallel equivalent resistance
- State Kirchhoff’s junction and loop rules and the conservation law behind each
- Solve a multiloop circuit with two EMFs for every branch current
EMF, internal resistance, and terminal voltage
A battery is characterized by its EMF ε — the energy per unit charge it supplies — and its internal resistance r. When the battery delivers current I, some voltage is lost inside it, so the terminal voltage actually available to the circuit is V = ε − Ir. Only with no current drawn (or an ideal battery, r = 0) does the terminal voltage equal the EMF.
Two rules, two conservation laws
When a circuit cannot be reduced to one loop, Kirchhoff’s rules take over. The junction rule — total current into a node equals total current out — is conservation of charge: charge cannot pile up at a junction. The loop rule — the potential changes around any closed loop sum to zero — is conservation of energy: a charge returning to its starting point must return to its starting potential. Every circuit, however tangled, yields to these two rules plus algebra.
A systematic recipe
Label a current in every branch with an assumed direction — a wrong guess is harmless, the algebra returns a negative value to tell you the true direction is opposite. Write the junction rule at enough nodes to relate the currents, then write one loop equation per independent loop. Solve simultaneously. Finally, sanity-check: currents into each node balance, and each loop equation closes to zero.
A battery with ε = 12 V and internal resistance r = 0.50 Ω drives a 5.5 Ω resistor. Find the current and the battery’s terminal voltage.
- 1.One loop: ε − Ir − IR = 0, so I = ε/(r + R).
- 2.I = 12 / (0.50 + 5.5) = 12 / 6.0 = 2.0 A.
- 3.Terminal voltage: V = ε − Ir = 12 − (2.0)(0.50) = 11 V.
- 4.Check: the external resistor gets V = IR = (2.0)(5.5) = 11 V — exactly the terminal voltage, as it must.
Two batteries and three resistors: branch 1 has ε₁ = 24 V in series with R₁ = 4.0 Ω, branch 2 is just R₂ = 4.0 Ω, and branch 3 has ε₂ = 6.0 V in series with R₃ = 4.0 Ω. All three branches connect the same two nodes, with both EMFs driving current toward the top node. Find the three branch currents.
- 1.Assume I₁ and I₃ flow up (toward the top node) through their batteries, and I₂ flows down through R₂. Junction rule at the top node: I₁ + I₃ = I₂.
- 2.Left loop (through ε₁ and R₂): 24 − 4.0·I₁ − 4.0·I₂ = 0, so I₁ + I₂ = 6.0.
- 3.Right loop (through ε₂ and R₂): 6.0 − 4.0·I₃ − 4.0·I₂ = 0, so I₃ + I₂ = 1.5.
- 4.Substitute I₃ = I₂ − I₁ into the second loop: (I₂ − I₁) + I₂ = 1.5 → 2I₂ − I₁ = 1.5. Add to I₁ + I₂ = 6.0: 3I₂ = 7.5 → I₂ = 2.5 A.
- 5.Then I₁ = 6.0 − 2.5 = 3.5 A and I₃ = 2.5 − 3.5 = −1.0 A. The minus sign means I₃ actually flows downward: the 24 V battery is recharging the 6 V battery.
- 6.Verify at the junction: 3.5 + (−1.0) = 2.5 A ✓, and both loop equations close to zero.
Loop-rule bookkeeping wins or loses the multiloop FRQ. Commit to this convention: pick a travel direction around each loop; a resistor traversed with its assumed current contributes −IR and against it +IR; a battery crossed from − to + contributes +ε, from + to − contributes −ε. State the junction equation explicitly — graders award a point for it even before any algebra.
Kirchhoff’s junction rule is a direct statement of which conservation law?
A battery with ε = 9.0 V and internal resistance 0.50 Ω delivers 2.0 A to a circuit. Its terminal voltage is:
Answer the 2 checkpoints as you read.
Sign in to save your progress