Circuit Calculations
Year 10 (IGCSE) ⚡ Electricity & Magnetism Solve circuit problems using V = IR, series/parallel rules.
🔗 Series Circuit Calculations
Solving series circuits requires applying Ohm's law and series rules systematically.
🔗 Series Circuit Rules
$$R_T = R_1 + R_2 + R_3 \qquad I = \frac{V_T}{R_T} \qquad V_n = I \times R_n$$🔌 Worked example: Two resistors 3Ω and 7Ω in series, connected to a 20 V supply.
R_T = 3 + 7 = 10 Ω
I = 20/10 = 2 A (same throughout)
V₁ = 2×3 = 6 V, V₂ = 2×7 = 14 V
Check: 6+14 = 20 V ✅
R_T = 3 + 7 = 10 Ω
I = 20/10 = 2 A (same throughout)
V₁ = 2×3 = 6 V, V₂ = 2×7 = 14 V
Check: 6+14 = 20 V ✅
⚡ Parallel Circuit Calculations
Parallel circuits need careful application of the 1/R_T formula.
⚡ Parallel Circuit Rules
$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} \qquad V = V_1 = V_2 \qquad I_T = I_1 + I_2$$⚡ Worked example: 6Ω and 12Ω in parallel, connected to 12 V.
1/R_T = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 → R_T = 4 Ω
I₁ = 12/6 = 2 A, I₂ = 12/12 = 1 A
I_T = 2+1 = 3 A. Check: V = I_T×R_T = 3×4 = 12 V ✅
1/R_T = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 → R_T = 4 Ω
I₁ = 12/6 = 2 A, I₂ = 12/12 = 1 A
I_T = 2+1 = 3 A. Check: V = I_T×R_T = 3×4 = 12 V ✅
🔋 Internal Resistance
Real batteries have internal resistance (r). When current flows, some voltage is "lost" inside the battery.
🔋 Terminal Voltage
$$V_{\text{terminal}} = \mathcal{E} - Ir$$
$\mathcal{E}$ = EMF (electromotive force, open-circuit voltage) · $r$ = internal resistance
🔋 Example: Battery with EMF 12 V, internal resistance 0.5 Ω, supplying 4 A.
V_terminal = 12 − (4×0.5) = 12 − 2 = 10 V
V_terminal = 12 − (4×0.5) = 12 − 2 = 10 V
When a car starts, the starter motor draws ~200 A. Even a small internal resistance of 0.01 Ω causes a voltage drop of 2 V — that's why your lights might dim briefly!
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🔌 Series & Parallel Circuit Solver
Ohm's Law: V = IR