Advanced Circuits · 14.1
Kirchhoff's Laws
Kirchhoff's two laws are conservation of charge and conservation of energy translated into circuit language — and they solve any network that the series and parallel rules cannot.
Build the model
Connect the measurement to the mechanism.
At any junction, the current in equals the current out, because charge cannot pile up or vanish. Around any closed loop, the e.m.f.s equal the sum of the potential drops, because each coulomb must spend exactly the energy it was given.
- Simple definition
- Kirchhoff's first law: currents into a junction sum to the currents out. Second law: around any loop, supplied e.m.f. equals the sum of potential drops.
- Example
- If 5 A enters a junction and 2 A leaves along one wire, the other wire must carry 3 A away — charge is conserved.
Whatever current arrives at a junction must leave it — charge cannot be created, destroyed, or stored at a wire joint.
Conservation of charge at every junction
Travel any complete loop and the energy given per coulomb by sources equals the energy dropped per coulomb across components.
Conservation of energy around every loop
The supply's voltage is shared out across the components of the loop — nothing is left over and nothing extra appears.
One equation per independent loop
Choose directions
Assign a direction to each unknown current. A negative answer simply means the true current flows the other way.
Write the equations
One junction equation per junction (minus one), one loop equation per independent loop. Cross a component against the current and the p.d. counts negative.
Solve simultaneously
Two unknowns need two equations; three need three. The laws always supply exactly enough.
Interactive circuit graph
Read the topology before the values.
Use this connectivity map alongside 14.1 Kirchhoff's Laws. Change the network, then select a node or branch to see what the graph preserves.
- Nodes
- Points where electrical connections meet.
- Branches
- Lines or general impedances connecting two nodes.
- Simplification
- Component details, such as specific resistor values, are hidden to focus purely on connectivity.
Bridge network: 4 nodes and 6 branches. Select a node or branch to investigate it.
Change one variable at a time
Make the relationship visible.
Whatever leaves through the branches must have entered from the supply — set the branches and the junction sets the input.
ΣI out2.5 A
ΣI in − ΣI out0 A — always
Catch the common trap
Explain before calculating.
A 12 V battery feeds three parallel resistors: 10 Ω, 100 Ω, 40 Ω. What total current does the battery supply?
Choose an answer to test the model.
Worked examples
State the rule, substitute, then check units.
EasyCurrents of 3.0 A and 2.0 A flow into a junction; one wire carries 4.0 A out. What does the remaining wire carry?
- ΣI in = ΣI out: 5.0 = 4.0 + I.
- I = 1.0 A out.
Answer1.0 A outward
MediumA 12 V supply drives two series resistors, 10 Ω and 30 Ω. Use the loop law to find the current and each p.d.
- Loop law: 12 = I × 10 + I × 30 = 40I.
- I = 0.30 A everywhere in the single loop (first law).
- V₁ = 3.0 V and V₂ = 9.0 V; check 3.0 + 9.0 = 12 V.
AnswerI = 0.30 A, with 3.0 V and 9.0 V across the resistors
HardA 9.0 V battery feeds two parallel branches: 3.0 Ω and 6.0 Ω. Use both of Kirchhoff's laws to find all three currents.
- Loop law per branch: I₁ = 9/3 = 3.0 A; I₂ = 9/6 = 1.5 A.
- Junction law: supply current = 3.0 + 1.5.
- I = 4.5 A.
Answer3.0 A, 1.5 A, and 4.5 A from the battery
ChallengingTwo batteries face each other: E₁ = 12 V (r = 1.0 Ω) and E₂ = 9.0 V (r = 0.5 Ω) drive a 4.5 Ω resistor in one loop, e.m.f.s opposing. Find the current and each battery's terminal p.d.
- Net e.m.f. = 12 − 9.0 = 3.0 V around the loop; total resistance = 1.0 + 0.5 + 4.5 = 6.0 Ω.
- I = 3.0 ÷ 6.0 = 0.50 A, driven by the stronger battery; the 9 V battery is being charged.
- V₁ = 12 − 0.5 × 1.0 = 11.5 V; V₂ = 9.0 + 0.5 × 0.5 = 9.25 V (charging raises its terminal p.d.).
AnswerI = 0.50 A; terminals read 11.5 V and 9.25 V