Series and parallel circuits
Real circuits combine series and parallel sections. You need three rules for each type of connection.
Series circuits — single loop
- Current is the same at every point: $I_1 = I_2 = I_3 = I_{\text{total}}$.
- Potential difference divides between components: $V_{\text{total}} = V_1 + V_2 + V_3$.
- Resistance adds: $R_{\text{total}} = R_1 + R_2 + R_3$.
Why? Charge can't accumulate (same I); battery supplies fixed energy per coulomb that gets shared (V splits); each resistor opposes flow, so opposition adds up (R adds).
Parallel circuits — branches
- Potential difference is the same across each branch: $V_1 = V_2 = V_{\text{total}}$.
- Current divides: $I_{\text{total}} = I_1 + I_2 + I_3$.
- Resistance is less than the smallest branch's resistance: $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ (HT only).
The parallel rule for two equal resistors: $R_{\text{total}} = R/2$. Two unequal: $R_{\text{total}} = \frac{R_1 R_2}{R_1 + R_2}$.
✦Worked example— Worked example — series
Two resistors of 4 Ω and 6 Ω in series with a 12 V cell.
- Total R = 4 + 6 = 10 Ω.
- Current I = V/R = 12/10 = 1.2 A.
- p.d. across 4 Ω: V₁ = IR = 1.2 × 4 = 4.8 V.
- p.d. across 6 Ω: V₂ = 1.2 × 6 = 7.2 V.
- Check: 4.8 + 7.2 = 12 V ✓.
✦Worked example— Worked example — parallel
Two 6 Ω resistors in parallel with a 12 V cell.
- p.d. across each = 12 V (same as supply).
- Each branch current = V/R = 12/6 = 2 A.
- Total current = 2 + 2 = 4 A.
- Equivalent resistance = V/I = 12/4 = 3 Ω (= 6/2 ✓).
Why parallel reduces resistance
Adding another branch creates an extra path for current. With more paths, more current flows for the same p.d. — total resistance falls. Mathematically, if you add a resistor in parallel, $R_{\text{total}}$ always decreases.
Required practical 4 — combining resistors
- Connect a fixed resistor R₁ to a battery; record I.
- Compute R from V/I.
- Add R₂ in series; predict and measure new total R.
- Re-connect R₂ in parallel; predict and measure new total R.
- Repeat for different combinations.
⚠Common mistakes
- Adding parallel resistances directly (R = R₁ + R₂) — wrong, must use reciprocals.
- Forgetting that p.d. is shared in series and shared currents in parallel.
- Drawing branches without proper junctions (no dot, no connection).
- Computing total parallel resistance and getting more than the smallest branch — sign that the maths went wrong.
➜Try this— Quick check
Three 6 Ω resistors in parallel: $\frac{1}{R} = \frac{3}{6} = \frac{1}{2}$ → R = 2 Ω.
AI-generated · claude-opus-4-7 · v3-deep-physics