Current, resistance & DC circuits
The idea
Current is the rate of charge flow, I = dq/dt, and resistance converts the microscopic picture — electrons drifting through a lattice, scattering as they go — into the circuit relation V = IR. University circuit analysis goes beyond series-parallel reduction to Kirchhoff's two rules, which are conservation laws in disguise: the junction rule (currents into a node sum to currents out) is conservation of charge, and the loop rule (potential changes around any closed loop sum to zero) is conservation of energy per charge.
The reliable procedure: assign a current with an assumed direction to each branch, write the junction rule at the nodes, then walk each independent loop adding potential changes — a drop of IR when you traverse a resistor with the current, a rise of the emf when you pass through a battery from − to +. Solve the linear system; a negative result simply means that current flows opposite to your guess, not that you made an error.
Many students believe a battery always pushes current out of its positive terminal. In multi-battery circuits a stronger source can force current backward through a weaker one — that is exactly what charging a battery means — so let the algebra decide the directions.
Worked example
Two batteries are connected between the same pair of nodes A and B: a 12 V battery in series with 2.0 Ω in one branch, a 6.0 V battery in series with 1.0 Ω in another, both positive terminals toward node A. A 4.0 Ω resistor forms the third branch between A and B. Find the three branch currents.
- Define unknowns: let I₁ flow from the 12 V branch into node A, I₂ flow from A to B through the 4.0 Ω resistor, and I₃ flow from A into the 6.0 V branch; the junction rule at A gives I₁ = I₂ + I₃.
- Walk the loop containing the 12 V battery and the 4.0 Ω resistor: +12 − 2.0 I₁ − 4.0 I₂ = 0, which with the junction rule becomes 12 = 2(I₂ + I₃) + 4I₂ = 6I₂ + 2I₃, i.e. 3I₂ + I₃ = 6.
- Walk the loop containing the 4.0 Ω resistor and the 6.0 V branch (from A down through the battery branch, back up through the resistor): −1.0 I₃ − 6.0 + 4.0 I₂ = 0, i.e. 4I₂ − I₃ = 6.
- Add the two loop equations to eliminate I₃: 7I₂ = 12, so I₂ = 12/7 ≈ 1.71 A; then I₃ = 4(12/7) − 6 = 6/7 ≈ 0.86 A and I₁ = I₂ + I₃ = 18/7 ≈ 2.57 A.
- Interpret I₃: it is positive flowing from A into the 6.0 V battery's positive terminal, so the 12 V source is recharging the 6.0 V battery rather than being assisted by it.
- Verify with node potentials: V_A = 12 − 2.0 × 2.57 ≈ 6.86 V, and the resistor branch gives 4.0 × 1.71 ≈ 6.86 V, while the battery branch gives 6.0 + 1.0 × 0.86 ≈ 6.86 V — all three branches agree.
Answer. I₁ ≈ 2.57 A from the 12 V branch, I₂ ≈ 1.71 A through the 4.0 Ω resistor, and I₃ ≈ 0.86 A charging the 6.0 V battery.
Check your understanding
- Which conservation law does each of Kirchhoff's rules express, and why must both hold simultaneously?
- What does a negative current in your solution actually tell you, and why is it not a mistake?
- Under what conditions does current flow into a battery's positive terminal, and what is happening energetically when it does?
- Why does series-parallel reduction fail for this circuit, forcing the full Kirchhoff treatment?
Build the foundations first
Current, resistance & DC circuits builds on these concepts. If any feel shaky, start there.