Apply Ohm's law, series and parallel rules and the power formulas to simple battery circuits.
Why this matters
Voltage is the energy each charge carries; current is how many charges go past per second; resistance is how hard the material makes them work to get through. Ohm's law ties these three together and is the first calculation in almost every circuit problem, from choosing a resistor for an LED to estimating how warm a wire gets. Electrical engineers, physicists and lab scientists use it daily, often without writing it down.
Current
Current I is the rate at which charge flows past a point:
I = Q / t units: amperes (A), where 1 A = 1 C/s
A current of 2 A moves 2 coulombs every second. Circuits often use milliamps: 20 mA = 0.020 A.
By convention, current is drawn flowing from the + terminal of a source, through the circuit, to the − terminal. (In metal wires the moving charges are actually electrons drifting the other way. The convention predates that discovery, and the maths works the same either way.)
Current only flows around a closed loop. Break the loop anywhere, with a switch or a broken wire, and current stops everywhere in that loop.
Resistance and Ohm's law
Resistance R, measured in ohms (Ω), describes how much voltage it takes to push a given current through a component. For many materials (metals, carbon film resistors) at a steady temperature, current is proportional to voltage. That is Ohm's law:
V = I · R I = V / R R = V / I
A 100 Ω resistor with 5 V across it carries I = 5 / 100 = 0.05 A = 50 mA.
Not everything obeys Ohm's law: diodes, LEDs and transistors do not, and a light bulb's filament resistance rises a lot as it heats up. For those, V = I·R still defines the resistance at one operating point, but R is not a constant.
What sets resistance
For a uniform wire:
R = ρ · L / A
ρ (rho) is the material's resistivity in Ω·m, L the length and A the cross-sectional area. Copper has ρ ≈ 1.68 × 10⁻⁸ Ω·m. So 10 m of copper wire with a 1.0 mm² (10⁻⁶ m²) cross-section has
R = 1.68 × 10⁻⁸ × 10 / 10⁻⁶ = 0.168 Ω
Longer wire means more resistance; thicker wire means less. That is why high-current cables are thick.
Series and parallel
Components in series share one path, so the same current flows through each, and their resistances add:
R_series = R₁ + R₂ + R₃ + ...
Components in parallel sit across the same two points, so they share the same voltage, and the currents add. The combined resistance is smaller than the smallest branch:
1 / R_parallel = 1 / R₁ + 1 / R₂ + ...
Two equal resistors in parallel give half the resistance of one.
Power
A current through a resistance turns electrical energy into heat at the rate
P = V · I = I² · R = V² / R (watts)
Every resistor has a power rating (commonly 0.25 W for small through-hole parts). Exceed it and the part overheats.
Worked example 1: a series circuit
A 9.0 V battery drives a 100 Ω and a 200 Ω resistor in series.
total resistance R = 100 + 200 = 300 Ω
current I = 9.0 / 300 = 0.030 A = 30 mA
voltage across 100 Ω: 0.030 × 100 = 3.0 V
voltage across 200 Ω: 0.030 × 200 = 6.0 V (3.0 + 6.0 = 9.0 V)
power in 100 Ω: 0.030² × 100 = 0.09 W
power in 200 Ω: 0.030² × 200 = 0.18 W
total power 9.0 × 0.030 = 0.27 W
The voltage divides in proportion to the resistances. That is a voltage divider, one of the most common sub-circuits in electronics.
Worked example 2: the same resistors in parallel
1 / R = 1/100 + 1/200 = 0.015 → R = 66.7 Ω
I from the battery = 9.0 / 66.7 = 0.135 A
Each branch has the full 9.0 V: 90 mA flows through the 100 Ω branch and 45 mA through the 200 Ω branch, 135 mA in total.
Worked example 3: an LED resistor
A red LED needs about 20 mA and drops about 2.0 V. From a 9.0 V battery, the series resistor must drop the other 7.0 V:
R = 7.0 / 0.020 = 350 Ω → use the next standard value up, 390 Ω
Rounding up keeps the current slightly below the maximum.
In code
def parallel(*resistances):
return 1 / sum(1 / r for r in resistances)
print(round(parallel(100, 200), 1)) # 66.7
Common mistakes
- Forgetting to convert mA. 30 mA is 0.030 A.
- Adding parallel resistances. Add reciprocals instead.
- Assuming the same current flows in parallel branches. Same voltage, different currents.
- Treating mains like a battery. Home wiring can kill. Practise on batteries of 9 V or less or in a simulator; real installation work needs training, supervision and, for electricians, a licence.
Quick recap
- I = Q / t, in amperes.
- V = I·R; R = ρL/A.
- Series resistances add; parallel reciprocals add.
- P = VI = I²R = V²/R.
Try it
In the Do tab, write the series-current function and use it to analyse a battery circuit.