Numford
Physics

Circuits

Build it. Watch the current flow.

Circuit builder

Tap a resistor to choose it, then change its value or add another in series or in parallel with it. The glowing dots are the current: faster where more flows. Brighter resistors turn more power into heat.

Equivalent resistance6 Ωbetween A and B
Current2 Afrom the cell
Power24 WP = VI
RIVPower
R₁4 Ω1.5 A6 V9 W
R₂12 Ω0.5 A6 V3 W
R₃3 Ω2 A6 V12 W

Step by step

  1. R₁ and R₂ in parallel: R = R₁R₂ / (R₁ + R₂) = 4 × 12 / (4 + 12) = 3 Ω
  2. The groups are in series: Req = 3 + 3 = 6 Ω
  3. Current from the cell: I = V / Req = 12 / 6 = 2 A
  4. Across the parallel group V = IR = 2 × 3 = 6 V, the same for every branch. It splits R₁: 6 / 4 = 1.5 A, R₂: 6 / 12 = 0.5 A.
  5. Drops in series: R₃: 2 × 3 = 6 V.
  6. Power: P = EI = 12 × 2 = 24 W.

Ohm's law

V = IR at a fixed temperature. Pick what to find and type the other two. Add a time to get the heat, H = I²Rt, and the energy in kWh.

Current I2 A
Power24 W
  1. I = V / R = 12 / 6 = 2 A
  2. Power: P = VI = 12 × 2 = 24 W. The same as I²R and V²/R.

Capacitors

In series the charge is the same on each and 1/C = 1/C₁ + 1/C₂ + 1/C₃. In parallel the voltage is the same and C = C₁ + C₂ + C₃. Energy U = ½CV². Leave a box empty to use fewer capacitors.

In series1 μF
Charge12 μC
Energy72 μJ
  1. C₁, C₂, C₃ in series: 1/C = 1/2 + 1/3 + 1/6 = 1, so C = 1 μF
  2. Charge from the cell: Q = CV = 1 × 12 = 12 μC
  3. In series every capacitor holds the same 12 μC, so V = Q/C splits the voltage: C₁: 6 V, C₂: 4 V, C₃: 2 V.
  4. Energy stored: U = ½CV² = ½ × 1 × 12² = 72 μJ

Resistivity

R = ρl/A. Resistivity ρ belongs to the material, at 20 °C; the bars use a log scale so copper and nichrome both show. Tap a material to use it.

Copper wire0.0162 Ωρ = 1.62 × 10⁻⁸ Ω m
  • Silver1.60 × 10⁻⁸0.016 Ω
  • Copper1.62 × 10⁻⁸0.0162 Ω
  • Aluminium2.63 × 10⁻⁸0.0263 Ω
  • Tungsten5.20 × 10⁻⁸0.052 Ω
  • Iron10.0 × 10⁻⁸0.1 Ω
  • Nichrome100 × 10⁻⁸1 Ω
  1. Copper: ρ = 1.62 × 10⁻⁸ Ω m
  2. A = 1 mm² = 1 × 10⁻⁶ m², since 1 mm² = 10⁻⁶ m²
  3. R = ρl / A = 1.62 × 10⁻⁸ × 1 / (1 × 10⁻⁶) = 0.0162 Ω

Kirchhoff's laws, two loops

Three branches join at the top and the bottom. A cell's + side faces up; type a negative emf to turn it round, or 0 for no cell. The junction rule and the two loop rules give three equations, solved exactly.

Branch 1
Branch 2
Branch 3
I₁2 Aup
I₂1.33 Adown
I₃0.67 Adown
  1. Take I₁, I₂ and I₃ flowing up each branch to the top junction. A negative answer means that current flows down.
  2. Junction rule at the top: I₁ + I₂ + I₃ = 0
  3. Left loop, clockwise: E₁ - I₁R₁ + I₂R₂ - E₂ = 0, so 12 - 2I₁ + 6I₂ = 0
  4. Right loop, clockwise: E₂ - I₂R₂ + I₃R₃ - E₃ = 0, so -6I₂ + 3I₃ - 6 = 0
  5. Solving the three equations: I₁ = 2 A, I₂ = -4/3 ≈ -1.33 A, I₃ = -2/3 ≈ -0.67 A
  6. 2 A up branch 1, 1.33 A down branch 2, 0.67 A down branch 3. Check: what flows up equals what flows down.
  7. The top junction is 8 V above the bottom: V = E₁ - I₁R₁ = 12 - 2 × 2 = 8 V

Wheatstone bridge

P from A to B, Q from B to C, R from A to D, S from D to C, the galvanometer G from B to D. It balances when P/Q = R/S; then no current flows through G.

BalancedIg = 0P/Q = R/S
S for balance30 Ω
Cell current0.33 A
  1. P/Q = 10 / 20 = 0.5 and R/S = 15 / 30 = 0.5.
  2. The ratios are equal, so B and D sit at the same potential and no current flows through the galvanometer.
  3. With G carrying nothing, P and Q are in series (10 + 20 = 30 Ω) and so are R and S (15 + 30 = 45 Ω). The two paths in parallel give 18 Ω between A and C.
  4. Cell current: I = E / R = 6 / 18 = 1/3 ≈ 0.33 A.
Worked examples

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Same subject: dimensional formula, unit converter, significant figures, error analysis, vectors, mirrors and lenses, projectile motion, atoms and quanta.

Questions students ask

Circuits, quickly

How do I find the equivalent resistance of resistors in series and parallel?

In series add them: Rs = R₁ + R₂ + R₃. In parallel add the reciprocals: 1/Rp = 1/R₁ + 1/R₂ + 1/R₃. For a mixed circuit replace each purely series or purely parallel group by its equivalent, and repeat until one resistor is left.

Why does the current split the way it does in parallel?

Every branch has the same voltage across it, so each takes I = V/R. The smaller the resistance, the larger its share. With 5, 10 and 30 ohm on 12 V the branches carry 2.4, 1.2 and 0.4 A, which add to the 4 A from the cell.

What do Kirchhoff's two laws say?

The junction rule: the currents into a junction add to the currents out of it (charge is conserved). The loop rule: round any closed loop the emfs add to the IR drops (energy is conserved). Together they solve any network.

When is a Wheatstone bridge balanced?

When P/Q = R/S. Then B and D are at the same potential and no current flows through the galvanometer, so any one resistance follows from the other three: S = QR/P.

Are the answers exact?

Yes. Every reduction, bridge and loop is worked out with exact fractions, so 5, 10 and 30 ohm in parallel give exactly 3 ohm and a current of 4/3 A is shown as 4/3, then rounded only for display.