Lesson 3 · 16 min
DC circuits
Charge, potential, resistance. Kirchhoff’s two rules are Newton and energy, in electrical clothes.
Ohm, then Kirchhoff
For many materials, V = IR over a useful range. R is geometry and resistivity: R = ρL/A. Long, thin, and poorly conducting makes a large R. Power dissipated in a resistor is I²R — that is why undersized cable runs hot.
Kirchhoff’s current law: charge is conserved, so currents into a node sum to zero. Kirchhoff’s voltage law: energy is conserved, so the signed sum of potential changes around a loop is zero. Together they size every DC network you will meet in a first circuits course.
Series shares current; parallel shares voltage. The equivalent of two equal resistors in parallel is half of one.
Worked example
A lamp and a heater
- Given: 12 V supply
- Given: lamp 6.0 Ω
- Given: heater 3.0 Ω, in parallel
Find: Supply current and total power
- R_p = (6 × 3) / (6 + 3) = 2.0 Ω
- I = V/R = 12/2 = 6.0 A
- P = IV = 12 × 6 = 72 W (lamp 24 W, heater 48 W)
6.0 A, 72 W. A 10 A fuse would hold; a 5 A fuse would not.
Check
Two 8 Ω resistors in parallel, across 16 V. Current through one of them?
Bench
Circuit
V = IR
- Req
- 2.00 Ω
- I supply
- 6.00 A
- I1 / I2
- 2.00 / 4.00 A
- Power
- 72.0 W