How to start a circuits problem

F1 · Node / mesh equations

Recognise it: "find the current / voltage", often with the method named. There are several sources and no switch.

If the method isn't named, count unknowns and pick the smaller.

The three lines:

1. Unknowns drawn on the figure: clockwise mesh currents, or node voltages with ground marked.
2. Constraints first. Write every line a source fixes (i₃ = −1 A, a supermesh difference). Then write every diamond's control variable in your unknowns (I_x = i₁). A dependent source is written like an ordinary source; it just adds this one extra line.
3. The first KVL/KCL, one element at a time as you walk round.

1. UnknownsThree clockwise mesh currents i₁, i₂, i₃, left to right.
2. ConstraintsThe 1 A source is on mesh 3's outer edge. Clockwise i₃ runs down that edge, but the arrow points up, so i₃ = −1 A. Control variables: I_x is the current in the top-left 1 Ω, which only mesh 1 passes through, so I_x = i₁. I_y flows down the right-hand inner 1 Ω, where i₂ runs down and i₃ runs up, so I_y = i₂ − i₃.
3. First KVLMesh 1, walking clockwise: up through the 5 V, right through 1 Ω, down through the I_y diamond from + to −, down through the shared 1 Ω: 5 − i₁ − I_y − (i₁ − i₂) = 0

Hand-over. Mesh 2's KVL is one more equation of the same kind. Substitute the two constraint lines and the algebra gives I_x = 2 A and I_y = 1 A. The 2026 paper's Q1 is this family with a supermesh added, and it paid [1M] per equation.

F2 · Thevenin / Norton

🎧 Watch it done first: Oct 2024 Q2, Norton with a dependent source — narrated, 6–8 min each, sound on. Each builds from the basic idea and stops to ask for your next move.

Recognise it: "equivalent at terminals a–b", "Norton's current source and resistance", or anything about maximum power. A maximum-power question is a Thevenin question first: R_L = R_th and P_max = V_th²/(4R_th).

The route to R_th is decided by one look at the circuit: are there diamonds?

The three lines: ① Remove the load, and label a (top) and b (ground). ② State the route and its reason: "dependent source present ⇒ V_oc/I_sc". ③ Write the open-circuit equations (KCL at each unknown node), with the control variable written in node voltages.

1. RouteThere is a 50 V source and a diamond, so the route is V_oc and I_sc, and R_N = V_oc/I_sc. The diamond is never switched off.
2. Unknownsb is ground. The control variable v_x is already a node voltage (the top of the 6 Ω), so the unknowns are v_x and V_a.
3. Open-circuit KCLWith currents leaving on the left and currents pushed in on the right: (v_x − 50)/3 + v_x/6 + (v_x − V_a)/2 = 0   and   (V_a − v_x)/2 + V_a/20 = 2.5v_x

Hand-over. Then short a–b (set V_a = 0) for I_sc. The answer comes out as I_N = 50 A and R_N = −20/19 Ω. The negative resistance is genuine: the diamond supplies power, so the network behaves like a negative resistor. The same route opens Oct 2024 Q2 (Norton with a VCVS 2V_a, 18 marks) and 2026 Q2(e), where the 32 V source becomes 2I.

F3 · Superposition

🎧 Watch it done first: Oct 2024 Q4, superposition with the dependent source kept on — narrated, 6–8 min each, sound on. Each builds from the basic idea and stops to ask for your next move.

Recognise it: "using the superposition theorem". It is always named, because it is almost never the fastest method, so the paper has to ask for it.

The three lines:

1. List the independent sources. That's how many partial circuits you draw. Dependent sources are not on the list, because they stay switched on in every partial circuit.
2. Redraw each partial circuit with the other independent sources off: a voltage source becomes a short (0 V is a wire) and a current source becomes an open circuit (0 A is a gap). The 2026 key gave [1M] for each of these drawings.
3. In each partial circuit, name the target with the same arrow or polarity as the original. Write its first equation, including the diamond's control line.

At the end, add the contributions with their signs. Never add powers: power isn't linear.

1. ListIndependent sources: 10 V, 8 A and 2 A, so three partial circuits. The 4i diamond is dependent: it stays on in all three.
2. Redraw10 V alone: 8 A → open, 2 A → open. 8 A alone: 10 V → short, 2 A → open. 2 A alone: 10 V → short, 8 A → open.
3. TargetIn each circuit, V₁ = 3i with i pointing down. The diamond line reads "top node minus the node below the diamond = 4i". Then write one KCL at the top node.

Hand-over. The three contributions are 30 V, −48 V and 12 V, so V₁ = −6 V. The minus sign means the lower end is positive.

F4 · Backwards from power

Recognise it: a power (or a limit on one) is given, and a source value is asked for.

The three lines: ① Start where the power is: P = I²R ⇒ I = √(P/R). ② Walk back one element at a time. An element in series carries the same current, so add its voltage drop. At a node, use KCL. ③ Keep walking until you reach the diamond's control variable or the source.

1. At the load18 = I_L² × 2 ⇒ I_L = 3 A, so the load voltage is 6 V.
2. Walk backThe 5 Ω next to the load carries the same 3 A, so the node above the middle 2 Ω sits at 6 + 5 × 3 = 21 V. The middle 2 Ω therefore takes 21/2 = 10.5 A.
3. Reach the controlKCL at that node: the 10 Ω brings in 10.5 + 3 = 13.5 A. That is the diamond's own current, so 2V_x = 13.5 ⇒ V_x = 6.75 V.

Hand-over. Now the left half: 6.75/5 = 1.35 A flows through the 5 Ω into the 2 Ω ∥ 4 Ω pair, so |V_o| = 6.75 + 1.35 × 4/3 = 8.55 V, and the source supplies 1.35 A. The paper asks for the magnitude because a power can't tell you the sign. The same first line opens 2023–24 MCQ 2 ("K such that the 2 Ω dissipates at most 50 W"): I = √(50/2) = 5 A.

F5 · First-order transient

🎧 Watch it done first: Mar 2025 Q2, RL with two switchings · Oct 2024 Q3(b), RC with a two-position switch — narrated, 6–8 min each, sound on. Each builds from the basic idea and stops to ask for your next move.

Recognise it: a switch at t = 0 (or "closed / opened for a long time"), with one L or C left once you've combined series and parallel pairs.

The six-line skeleton. The first three lines are the opening, and each is a separate small picture:

1. t < 0 pictureA long time before the switch, the capacitor is an open circuit and the inductor a short. Find v_C(0⁻) or i_L(0⁻).
2. Continuityv_C(0⁺) = v_C(0⁻) and i_L(0⁺) = i_L(0⁻). Nothing else is guaranteed to carry over, and resistor currents can jump.
   Unpack this step

   The stored energy is ½Cv² or ½Li², and energy can't change in zero time. That would need infinite power. So v_C and i_L are continuous. The Module 3 lesson has the full picture.

3. t → ∞ pictureAfter the switch, a long time later, the same rule (C open, L short) gives the final value x(∞).
4. R seen by L or CSwitch off the independent sources, keep any diamonds, and find the resistance at the element's terminals (with a test source if there's a diamond).
5. ττ = R_thC or L/R_th.
6. Formulax(t) = x(∞) + [x(0⁺) − x(∞)]e^−t/τ. Any other quantity follows from x(t).

Endings the papers bolt on:

• A second switch at t₀. Restart the clock at t₀ with the new picture: x(∞₂) + [x(t₀) − x(∞₂)]e^{−(t−t₀)/τ₂}.
• "How long until it reaches V₁". Set x(t) = V₁ and take ln.
• "Energy absorbed in (0, ∞)". Integrate ∫i²R dt.
• "i, di/dt, d²i/dt² at 0⁺". Write the KVL for t > 0 and evaluate it at 0⁺. Then differentiate the KVL and evaluate again (2023–24, subjective Q2).
• "Energy stored". The answer is line 1 alone, with C open and L short.

1. Two intervalsThere are two switching events, so the answer comes in two pieces, 0 < t < 4 s and t > 4 s. Each has its own start value, final value and τ.
2. Interval 1: start and endS₁ was open, so no source reached the inductor: i(0⁻) = 0 = i(0⁺). With S₂ still open and L a short at ∞, i(∞) = 40/(8 + 12) = 2 A.
3. Interval 1: τWith the 40 V shorted, the inductor sees 8 + 12 = 20 Ω, so τ₁ = 10/20 = 0.5 s and i = 2(1 − e^−2t) A.

Hand-over. At t = 4 s, run the same three lines again, starting from i(4) = 1.9993 A. The Thevenin equivalent at P is 20 V behind 8/3 Ω, so i(∞) = 15/11 A and τ₂ = 15/22 s. The two asked-for values are i(2) = 1.963 A and i(5) = 1.510 A.

F6 · Second-order transient

🎧 Watch it done first: Oct 2024 Q5(b), series RLC · Mar 2025 Q3, parallel RLC — narrated, 6–8 min each, sound on. Each builds from the basic idea and stops to ask for your next move.

Recognise it: after the switch, an L and a C are both in the circuit. Or the question uses the words: damping, α, ω₀, overdamped, natural frequency, damping ratio.

The three lines:

1. t < 0 picture. Find v_C(0⁻) and i_L(0⁻) (C open, L short). Both carry over to 0⁺.
2. The derivative at 0⁺, from the circuit. Write one KCL or KVL at 0⁺ using the two known values, then read off dv_C/dt(0⁺) = i_C(0⁺)/C or di_L/dt(0⁺) = v_L(0⁺)/L.
   Unpack this step

   It is just the element law rearranged. i_C = C dv_C/dt gives dv_C/dt = i_C/C. Likewise v_L = L di_L/dt gives di_L/dt = v_L/L. At 0⁺ you know every capacitor voltage and every inductor current, so i_C and v_L come from ordinary KCL and KVL.

3. α and ω₀ → case → form. For a series RLC, α = R/(2L). For a parallel RLC, α = 1/(2RC). Both have ω₀ = 1/√(LC). If the question asks for the damping ratio, it is ζ = α/ω₀ (2026 Q4(e)).

If a DC source is still connected after the switch, add the forced term: the final value from the t → ∞ picture. Then fit A and B to the complete response.

1. t < 0 pictureIn position 1 for a long time, C is open, so no current flows in the 8 Ω and v(0⁻) = 4 V. The L ∥ 0.5 Ω pair has no source, so i_L(0⁻) = 0. Both carry over to 0⁺.
2. Derivative at 0⁺In position 2, C, 0.5 Ω and L are in parallel. KCL at the top node, summing currents leaving: 1 · dv/dt + v/0.5 + i_L = 0. At 0⁺ this gives dv/dt(0⁺) = −(4/0.5 + 0)/1 = −8 V/s.
3. CaseParallel RLC: α = 1/(2 × 0.5 × 1) = 1 s⁻¹ and ω₀ = 1/√(0.25 × 1) = 2 rad/s. Since α < ω₀ it is underdamped, and v = e^−t(A cos √3t + B sin √3t). There is no forced term, because no source is connected after the switch.

Hand-over. From v(0⁺), A = 4. From the derivative, −A + √3B = −8, so B = −4/√3 ≈ −2.309. The derivation question (Oct 2024 Q5(a), 6 marks) opens with the series-loop KVL: L di/dt + Ri + v_C = 0 with i = C dv_C/dt. That gives the second-order ODE, then the characteristic equation, then complex roots when α < ω₀. The Module 3 lesson walks through it.

F7 · AC phasors scope to confirm

Recognise it: j in the element values, sinusoidal sources, "phasor", "average power", "peak values, not RMS".

The three lines:

1. Convert everything to phasors. Z_L = jωL and Z_C = −j/(ωC). A source V_m cos(ωt + φ) becomes V_m∠φ. Note whether the values are peak or RMS.
2. Use the DC method you'd use anyway (nodal, mesh or Thevenin), with complex numbers.
3. Power from peak phasors: P = ½ Re(VI*). For a resistor, that is ½|I|²R.

1. PhasorsThe values are already phasors (j1.92 Ω, −j2 A). The amplitudes are peak, so each resistor absorbs ½|V|²/R.
2. UnknownsNode voltages V₁ (at N₁) and V₂ (at N₂). The control variable is I_x = (V₂ − V₁)/(j1.92).
3. KCLAt N₂: 1.6I_x = I_x + V₂/8. At N₁: −j2 + I_x = V₁/4.8.

Hand-over. Solving gives I_x = 5 A, V₂ = 24 V and V₁ = 24 − j9.6 V. The 4.8 Ω absorbs 69.6 W, the 8 Ω absorbs 36 W, and the inductor absorbs 0 W. Parts (c) and (d), as printed, are flawed: the dependent source supplies power, so the remaining network's Thevenin impedance has a negative real part. The past-papers page discusses it.