Electric Circuits Problem Solving

This final lesson brings together everything you have learned about electric circuits. The key to success in A-Level circuit questions is a systematic approach: identify the circuit structure, choose the right equations, and work through the calculation step by step.

Strategy for Multi-Step Circuit Problems

When faced with a complex circuit problem, follow this approach:

flowchart TD
    A[Read the question and draw/annotate the circuit] --> B[Label all known values: EMF, R, I, V]
    B --> C[Identify series and parallel sections]
    C --> D[Simplify parallel combinations into equivalent R]
    D --> E[Find R_total including internal resistance if given]
    E --> F[Calculate total current: I = ε / R_total]
    F --> G[Work outwards: find V and I for each section]
    G --> H{Verification checks}
    H --> I[Do voltages around each loop sum to EMF?]
    H --> J[Do currents at each junction balance?]
    H --> K[Is R_parallel less than smallest R?]
    H --> L[Are units correct and significant figures appropriate?]

Worked Example 1: Combined Series-Parallel Circuit

A 24 V battery with negligible internal resistance is connected to the following arrangement: a 6.0 Ω resistor is in series with a parallel combination of 4.0 Ω and 12 Ω resistors. Find (a) the total resistance, (b) the current from the battery, (c) the p.d. across each section, and (d) the current through each parallel branch.

Solution:

(a) Parallel combination: R_p = (4.0 × 12)/(4.0 + 12) = 48/16 = 3.0 Ω Total resistance: R_total = 6.0 + 3.0 = 9.0 Ω

(b) Total current: I = V/R = 24/9.0 = 2.67 A

(c) P.d. across 6.0 Ω: V = IR = 2.67 × 6.0 = 16.0 V P.d. across parallel section: V = IR = 2.67 × 3.0 = 8.0 V Check: 16.0 + 8.0 = 24 V ✓

(d) Current through 4.0 Ω: I = V/R = 8.0/4.0 = 2.0 A Current through 12 Ω: I = V/R = 8.0/12 = 0.67 A Check: 2.0 + 0.67 = 2.67 A ✓

Worked Example 2: EMF and Internal Resistance with Parallel Load

A battery with EMF 9.0 V and internal resistance 1.5 Ω is connected to two resistors: 4.5 Ω and 6.0 Ω in parallel as the external load. Calculate (a) the external resistance, (b) the current from the battery, (c) the terminal p.d., and (d) the power dissipated in the 6.0 Ω resistor.

Solution:

(a) External resistance: R_ext = (4.5 × 6.0)/(4.5 + 6.0) = 27/10.5 = 2.57 Ω

(b) Total circuit resistance = R_ext + r = 2.57 + 1.5 = 4.07 Ω Current: I = ε/(R_ext + r) = 9.0/4.07 = 2.21 A

(c) Terminal p.d. = ε - Ir = 9.0 - (2.21 × 1.5) = 9.0 - 3.32 = 5.68 V (This is also the p.d. across both parallel resistors.)

(d) Power in 6.0 Ω: P = V²/R = (5.68)²/6.0 = 32.3/6.0 = 5.4 W

Worked Example 3: Potential Divider with Internal Resistance

A cell with EMF 6.0 V and internal resistance 2.0 Ω is connected to a potential divider made of a 4.0 Ω and an 8.0 Ω resistor. Calculate the output voltage across the 8.0 Ω resistor.

Solution:

Total resistance = r + R₁ + R₂ = 2.0 + 4.0 + 8.0 = 14 Ω Current: I = ε/R_total = 6.0/14 = 0.4286 A V_out = IR₂ = 0.4286 × 8.0 = 3.43 V

Note: without internal resistance, V_out would be 6.0 × 8.0/(4.0 + 8.0) = 4.0 V. The internal resistance reduces the output voltage. This is a common exam question — they give you a potential divider with a real battery and expect you to include r.

Worked Example 4: Potential Divider with a Sensor

A fire alarm circuit uses a potential divider with a fixed 10 kΩ resistor and an NTC thermistor. The supply is 9.0 V. The thermistor has a resistance of 40 kΩ at 20°C and 2.0 kΩ at 100°C. The thermistor is R₁ (top) and the fixed resistor is R₂ (bottom). The alarm triggers when V_out exceeds 7.0 V. Will the alarm trigger at 100°C?

Solution:

At 100°C: V_out = V_in × R₂/(R₁ + R₂) = 9.0 × 10/(2.0 + 10) = 9.0 × 10/12 = 7.5 V

Since 7.5 V > 7.0 V, yes, the alarm will trigger at 100°C.

At 20°C: V_out = 9.0 × 10/(40 + 10) = 9.0 × 10/50 = 1.8 V (well below threshold).

Worked Example 5: Kirchhoff's Laws Application

Two batteries are connected as follows: a 12 V battery (internal resistance 1.0 Ω) and a 6.0 V battery (internal resistance 0.50 Ω) are connected in opposition across an 8.0 Ω resistor. Find the current.

Solution:

The batteries oppose each other, so the net EMF = 12 - 6.0 = 6.0 V. Total resistance = R + r₁ + r₂ = 8.0 + 1.0 + 0.50 = 9.5 Ω Current: I = 6.0/9.5 = 0.63 A

The current flows in the direction driven by the larger (12 V) battery. The 6.0 V battery is being charged.

Worked Example 6: Complete Circuit Analysis with Power

A 15 V battery (r = 1.0 Ω) is connected to a 2.0 Ω resistor in series with a parallel combination of 6.0 Ω and 3.0 Ω. Calculate every current, every voltage, and every power dissipation.

Solution:

Step 1 — Simplify:

R_parallel = (6.0 × 3.0)/(6.0 + 3.0) = 18/9.0 = 2.0 Ω
R_total = r + R_series + R_parallel = 1.0 + 2.0 + 2.0 = 5.0 Ω

Step 2 — Total current:

I = ε/R_total = 15/5.0 = 3.0 A

Step 3 — Voltages:

Lost volts = Ir = 3.0 × 1.0 = 3.0 V
V across 2.0 Ω = IR = 3.0 × 2.0 = 6.0 V
V across parallel section = IR = 3.0 × 2.0 = 6.0 V
Check: 3.0 + 6.0 + 6.0 = 15 V ✓

Step 4 — Branch currents:

I through 6.0 Ω = V/R = 6.0/6.0 = 1.0 A
I through 3.0 Ω = V/R = 6.0/3.0 = 2.0 A
Check: 1.0 + 2.0 = 3.0 A ✓

Step 5 — Power dissipation:

Component	V (V)	I (A)	P (W)	Formula Used
Internal resistance	3.0	3.0	9.0	P = I²r
2.0 Ω series	6.0	3.0	18	P = I²R
6.0 Ω parallel	6.0	1.0	6.0	P = V²/R
3.0 Ω parallel	6.0	2.0	12	P = V²/R
Total	—	—	45 W	—

Check: Total power = εI = 15 × 3.0 = 45 W ✓

Worked Example 7: Loaded Potential Divider with Real Battery

A battery (ε = 10 V, r = 2.0 Ω) drives a potential divider: R₁ = 6.0 Ω (top) and R₂ = 12 Ω (bottom). A 12 Ω load is connected across R₂. Find V_out.

Solution:

Step 1 — Find combined bottom resistance:

R₂ ∥ R_L = (12 × 12)/(12 + 12) = 144/24 = 6.0 Ω

Step 2 — Total resistance:

R_total = r + R₁ + R_combined = 2.0 + 6.0 + 6.0 = 14 Ω

Step 3 — Current:

I = ε/R_total = 10/14 = 0.714 A

Step 4 — V_out:

V_out = I × R_combined = 0.714 × 6.0 = 4.3 V

Compare: Without load and without internal resistance, V_out = 10 × 12/(6.0 + 12) = 6.67 V. Both internal resistance and loading reduce the output.

Exam Technique Tips

Show Your Working

In A-Level exams, method marks are often worth more than the final answer. Always:

Write the formula you are using.
Substitute in the values with units.
Show intermediate steps.
Give your final answer with the correct unit and appropriate significant figures.

Check Your Answers

Do voltages in a series loop add up to the EMF?
Do currents at a junction balance?
Is the total resistance sensible? (Parallel combination should be less than the smallest resistor.)
Are the units correct?

Common Pitfalls

Pitfall	How to Avoid It
Forgetting to convert units	kΩ → Ω, mm → m, minutes → seconds. Do conversions first.
Using the wrong power formula	Check: what quantities do you actually know? Use the matching formula.
Ignoring internal resistance	If the question mentions EMF and r, you must include r in R_total.
Not reading the question	Check: do they want current, voltage, resistance, power, or energy?
Rounding too early	Keep 3+ significant figures in intermediate steps. Only round the final answer.
Confusing EMF with terminal p.d.	ε = total energy per coulomb; V = ε − Ir = voltage across external circuit.
Forgetting to take reciprocal in parallel	1/R_total = 1/R₁ + 1/R₂ → R_total = ... (do the final step!)

Summary of Key Equations

Quantity	Equations	Notes
Charge	Q = It	I in A, t in s
Potential difference	V = W/Q	W in J, Q in C
Resistance	R = V/I	Definition — always valid
Resistivity	R = ρL/A	ρ in Ω m, L in m, A in m²
Series resistance	R_total = R₁ + R₂ + ...	Always larger than any single R
Parallel resistance	1/R_total = 1/R₁ + 1/R₂ + ...	Always smaller than any single R
Two in parallel	R_total = R₁R₂/(R₁ + R₂)	Quick shortcut for two resistors
EMF / internal resistance	ε = V + Ir = I(R + r)	V is terminal p.d.
Potential divider	V_out = V_in × R₂/(R₁ + R₂)	Only valid with no load current
Power	P = IV = I²R = V²/R	Three equivalent forms
Energy	E = Pt = IVt	t in seconds for J, in hours for kWh
Kirchhoff's 1st law	ΣI_in = ΣI_out	Conservation of charge
Kirchhoff's 2nd law	Σε = ΣIR	Conservation of energy

A-Level Deep Dive: Electric Circuits Problem Solving

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, synoptic across sub-topics 3.1 to 3.5, requires candidates to bring together charge, current and potential difference (3.1), Ohm's law and I–V characteristics (3.2), resistivity (3.3), series and parallel networks (3.4), and emf with internal resistance (3.5), and to apply Kirchhoff's first and second laws together with the potential-divider relation in unfamiliar mixed networks (refer to the official Pearson Edexcel specification document for exact wording). Problem-solving questions in this strand are examined in 9PH0-01 (Paper 1, Advanced Physics I) and especially in 9PH0-03 (Paper 3, General and Practical Principles in Physics), where multi-step circuit analysis is combined with experimental data and graph interpretation. The Edexcel formula booklet provides $V = IR$ , $P = IV = I^2R = V^2/R$ , $\varepsilon = I(R + r)$ and the series/parallel resistor combination rules, but does not state Kirchhoff's laws or the potential-divider equation in symbolic form — these must be applied from understanding.

Worked example with full mark scheme

Question (12 marks):

The circuit shown contains two cells. Cell X has emf $\varepsilon_X = 6.0\ \text{V}$ and internal resistance $r_X = 0.50\ \Omega$ . Cell Y has emf $\varepsilon_Y = 4.0\ \text{V}$ and internal resistance $r_Y = 1.0\ \Omega$ . The two cells are connected in parallel (positive terminal to positive terminal). The combined source is connected across an external network consisting of a $3.0\ \Omega$ resistor in series with a parallel combination of a $6.0\ \Omega$ resistor and a $12\ \Omega$ resistor.

(a) Calculate the resistance of the external network. (2)

(b) Use Kirchhoff's laws to set up the simultaneous equations for the currents $I_X$ (from cell X), $I_Y$ (from cell Y), and $I$ (through the external network). (3)

(d) Calculate the power dissipated in the $12\ \Omega$ resistor and comment on whether cell Y is delivering or absorbing energy. (3)

Solution with mark scheme:

(a) Step 1 — combine the parallel pair.

$\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{12} = \dfrac{2 + 1}{12} = \dfrac{3}{12} \implies R_p = 4.0\ \Omega$

Step 2 — add the series resistor.

$R_\text{ext} = 3.0 + 4.0 = 7.0\ \Omega$ .

M1 — correct parallel-combination calculation. A1 — $R_\text{ext} = 7.0\ \Omega$ .

(b) Apply Kirchhoff's first law at the junction where both cells feed the external network: $I = I_X + I_Y$ (taking both cells to drive current outward through their positive terminals).

B1 — junction equation $I = I_X + I_Y$ .

Apply Kirchhoff's second law around the loop containing cell X and the external network:

$\varepsilon_X = I_X r_X + I R_\text{ext} \implies 6.0 = 0.50 I_X + 7.0 I$

M1 — loop equation for cell X with internal-resistance term and external-network term.

Apply Kirchhoff's second law around the loop containing cell Y and the external network:

$\varepsilon_Y = I_Y r_Y + I R_\text{ext} \implies 4.0 = 1.0 I_Y + 7.0 I$

M1 — loop equation for cell Y, sign-consistent with the chosen junction directions.

From cell X: $I_X = (6.0 - 7.0 I) / 0.50 = 12 - 14 I$ .

From cell Y: $I_Y = (4.0 - 7.0 I) / 1.0 = 4.0 - 7.0 I$ .

M1 — algebraic rearrangement of both loop equations for the cell currents.

Step 2 — substitute into the junction equation.

$I = I_X + I_Y = (12 - 14 I) + (4.0 - 7.0 I) = 16 - 21 I$

M1 — substitution into junction equation.

Step 3 — solve.

$I + 21 I = 16 \implies 22 I = 16 \implies I \approx 0.727\ \text{A}$ .

Then $I_X = 12 - 14(0.727) \approx 12 - 10.18 \approx 1.82\ \text{A}$ and $I_Y = 4.0 - 7.0(0.727) \approx 4.0 - 5.09 \approx -1.09\ \text{A}$ .

A1 — $I \approx 0.73\ \text{A}$ . A1 — $I_X \approx 1.8\ \text{A}$ and $I_Y \approx -1.1\ \text{A}$ (negative sign retained and interpreted).

(d) The two parallel resistors share the same potential difference. The current through the parallel pair is $I = 0.727\ \text{A}$ , dropped across $R_p = 4.0\ \Omega$ , giving $V_p = I R_p = 0.727 \times 4.0 \approx 2.91\ \text{V}$ .

Power in the $12\ \Omega$ resistor: $P_{12} = V_p^2 / 12 = (2.91)^2 / 12 \approx 0.71\ \text{W}$ .

M1 — correct identification that the parallel pair shares $V_p$ and use of $P = V^2/R$ on the $12\ \Omega$ branch.

A1 — $P_{12} \approx 0.71\ \text{W}$ .

B1 — cell Y has negative current $I_Y$ , meaning conventional current flows into its positive terminal: cell Y is being charged (absorbing energy) by cell X, which is the dominant source.

Total: 12 marks (M1 A1 + B1 M1 M1 + M1 M1 A1 A1 + M1 A1 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (8 marks): A potential divider is constructed from a fixed $1.0\ \text{k}\Omega$ resistor in series with a thermistor whose resistance varies from $5.0\ \text{k}\Omega$ at $0\ ^\circ\text{C}$ to $0.20\ \text{k}\Omega$ at $80\ ^\circ\text{C}$ . The divider is supplied by a $6.0\ \text{V}$ source of negligible internal resistance. The output is taken across the fixed resistor.

(a) Calculate the output voltage at $0\ ^\circ\text{C}$ and at $80\ ^\circ\text{C}$ . (3)

Strategy for Multi-Step Circuit Problems

Worked Example 1: Combined Series-Parallel Circuit

Worked Example 2: EMF and Internal Resistance with Parallel Load

Worked Example 3: Potential Divider with Internal Resistance

Worked Example 4: Potential Divider with a Sensor

Worked Example 5: Kirchhoff's Laws Application

Worked Example 6: Complete Circuit Analysis with Power

Worked Example 7: Loaded Potential Divider with Real Battery

Exam Technique Tips

Show Your Working

Check Your Answers

Common Pitfalls

Summary of Key Equations

A-Level Deep Dive: Electric Circuits Problem Solving

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

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