How to answer an electrical circuits viva question
A circuits viva tests more than whether you can reach a number. You need to explain the model you chose, state the assumptions behind it, show enough working to make your result checkable, and interpret what the result means. Practise speaking in that order rather than reciting equations without context.
For almost any circuit problem, use this sequence:
- Clarify the circuit and the quantity requested. Identify the reference node, source values, component connections and whether values are peak or RMS.
- Name the method. For example, “I’ll use nodal analysis and Kirchhoff’s current law at the unknown node.”
- State assumptions and sign convention. Say which direction you have assumed for current, whether a source is ideal, and what initial condition applies.
- Work through the key equation aloud. Keep units with values and distinguish a complex impedance from its magnitude.
- Check and interpret the result. Use a limiting case, conservation of current or power, and explain what the sign or phase tells you.
If you make an algebra error, do not conceal it by jumping to a different answer. Identify the step, correct it, then check whether the corrected value is physically plausible. In practice, concise and traceable reasoning is more useful than a long unstructured calculation.
The examiner strip in MySummaries is set up here as an engineering viva practice persona, not as a claim about a particular institution’s assessment rules.
A station list to practise from
Start with a small set of stations that test different kinds of reasoning. For each one, give yourself a short, uninterrupted answer before looking at a worked response. Then repeat the station and try to make your explanation clearer, not merely faster. These opening prompts are deliberately specific: they require you to set up and explain a circuit, not just name a formula.
The first three stations below are worked examples. Try speaking each answer before reading the model. A useful target is a clear explanation that includes the requested result, not a memorised speech. If you cannot finish a calculation, say what equation you would use and what information you still need.
Station 1: A loaded resistive divider
This station checks whether you recognise that the two lower resistors form a parallel load, and whether your nodal equation accounts for all current leaving the node. State the ground reference and current directions before calculating. The figures below show one practice attempt and feedback; the score is the platform’s feedback, not a universal viva grade.
Examiner
A 12 V ideal source feeds a 2 kΩ resistor in series with a node connected to 4 kΩ and 6 kΩ resistors in parallel to ground. Find the node voltage and both branch currents, and explain your method.
You identified the reference node and treated the source as ideal.
ImproveSay explicitly that the 2 kΩ resistor connects the 12 V node to the unknown node.
You combined the parallel resistance correctly and found a consistent node voltage.
ImproveShow the conductance form of KCL so both branch currents are visibly included.
The voltage and branch currents have appropriate units and signs.
ImproveCheck that the two branch currents sum to the current through the 2 kΩ resistor.
A strong answerI will take the bottom rail as ground and call the unknown node voltage V. The 12 V source is ideal, so KCL at the node gives (12 − V)/2 kΩ = V/4 kΩ + V/6 kΩ. The right-hand conductance is 1/4 kΩ + 1/6 kΩ, so V = 12 × (2.4/4.4) = 6.55 V, to three significant figures. The 4 kΩ branch current is 6.55/4 kΩ = 1.64 mA, and the 6 kΩ branch current is 6.55/6 kΩ = 1.09 mA. Their sum, 2.73 mA, matches (12 − 6.55)/2 kΩ, which checks KCL.
That is a MySummaries station, filled with electrical circuits material. Yours is written from your own notes. Start free
Notice how the check is part of the answer rather than an afterthought. A common weak response says “the divider gives about 6.5 volts” without explaining that the parallel branches load the source through the 2 kΩ resistor. If you are asked for one branch current, calculate it from that branch’s own resistance and the node voltage; do not split the source current equally unless the branch resistances are equal.
Station 2: A capacitor after a voltage step
Here the main test is whether you connect the time constant to the circuit’s exponential response and use the initial condition. Say what the capacitor voltage can and cannot do at the instant of switching: for an initially uncharged ideal capacitor, its voltage cannot jump instantaneously.
Examiner
An initially uncharged 100 μF capacitor is connected to an ideal 10 V step through a 10 kΩ resistor. Describe the capacitor voltage over time and find its value at one time constant.
You correctly gave 0 V initially and 10 V as the steady-state value.
ImproveMake clear that the capacitor voltage is continuous at the switching instant.
You used the charging exponential and obtained the correct value at one time constant.
ImproveShow the product R C with units before substituting into the exponential.
You described charging current decreasing as the capacitor voltage rises.
ImproveMention that the resistor voltage and current both approach zero at steady state.
A strong answerThe time constant is τ = R C = 10 kΩ × 100 μF = 1.0 s. The capacitor starts at 0 V and approaches the 10 V source value, so for t ≥ 0 its voltage is vC(t) = 10(1 − e^(−t/τ)) V. At one time constant, vC(τ) = 10(1 − e^(−1)) = 6.32 V. The resistor current is initially 10 V/10 kΩ = 1.0 mA and then decays as the capacitor charges. At steady state the capacitor voltage is 10 V and the current through the resistor is zero.
A frequent mistake is to say the capacitor is “fully charged after one time constant”. At one time constant it has reached about 63.2% of the total voltage change; it approaches its final value asymptotically. Also check the circuit before using τ = RC: in a circuit with more components, the resistance relevant to the time constant is the resistance seen by the capacitor with independent sources deactivated, not necessarily one resistor chosen by inspection.
Station 3: A series RC circuit in sinusoidal steady state
This station distinguishes candidates who can handle complex impedance from those who use only magnitudes. Keep the source’s RMS convention, the capacitor’s negative imaginary impedance, and the phase of current consistent throughout.
Examiner
A 10 V RMS source at 50 Hz drives a series circuit of 30 Ω and 100 μF. Find the impedance and current, then explain the power factor.
You used the correct capacitive reactance and included its negative imaginary sign.
ImproveState the angular frequency as 2πf before evaluating the reactance.
You found a current magnitude close to 0.229 A.
ImproveExplain that current leads the source voltage in this series capacitive circuit.
You identified a leading power factor and used the resistive component for real power.
ImproveGive the approximate power factor and make clear that the 10 V value is RMS.
A strong answerThe angular frequency is ω = 2πf = 314.2 rad/s, so the capacitor impedance is 1/(jωC) = −j31.8 Ω. The total impedance is therefore 30 − j31.8 Ω, with magnitude about 43.8 Ω and angle about −46.7°. The RMS current is 10/(30 − j31.8) = 0.229∠46.7° A, so current leads the source voltage. The power factor is cos(46.7°), about 0.685 leading. Real power is dissipated in the resistor: I²R ≈ (0.229)² × 30 = 1.57 W.
The most important correction is the phase relationship: a negative impedance angle gives a positive current angle when the source voltage is the 0° reference. Calling the power factor “0.685 lagging” would contradict the capacitive circuit. Another check is real power: an ideal capacitor stores and returns energy, while the resistor accounts for the average real power dissipated here.
Listen for the mark-losing phrase
When you review a recording, identify the exact phrase that lost useful detail. “The current leads” is better than saying nothing, but it is stronger to connect the lead to the negative capacitive reactance and the calculated phase angle. Feedback should point to language you can replace on the next attempt.
I use the RMS source phasor as the 0 degree reference, and the capacitor impedance is negative imaginary. So the current is about 0.229 amps. The current leads the voltage because the total impedance angle is negative. The power factor is about 0.685, leading, and the resistor dissipates the real power.
So the current is about 0.229 amps.
The magnitude is useful but does not answer the phase part of the question, and the unit alone does not say whether this is RMS or peak current.
Say: Say: ‘The RMS current is 0.229∠46.7° A, so it leads the source voltage by about 46.7°.’A short examiner-style debrief can help you hear whether you included the key interpretation, rather than only the arithmetic.
You obtained the current magnitude, but stating its phase would make your capacitive-circuit reasoning complete.
Turn practice into a repeatable method
Use the same routine for the remaining topics, including Thevenin equivalents, source transformations, mesh analysis and power calculations:
- Before answering: redraw or describe the circuit in words, mark the reference node and identify the requested quantity.
- During the answer: state the law or equivalent circuit you are using, show the decisive equation, and keep units and signs visible.
- At the end: check conservation, a limiting case or a physical expectation. For a Thevenin equivalent, for example, find the open-circuit voltage and then deactivate independent sources to find the resistance seen from the terminals.
- After recording: listen once for technical accuracy and once for communication. Note one knowledge gap and one phrase that was too vague, then repeat the station.
For a weekly routine, alternate calculation-heavy and explanation-heavy stations. A short session might be: one attempt without notes, a review of the step where your reasoning became unclear, and a second attempt using the corrected explanation. Keep a record of recurring errors—sign conventions, RMS versus peak values, initial conditions, or omitted checks—so the next session targets an actual weakness.
How MySummaries helps
MySummaries can turn your electrical circuits revision board into viva stations, record your spoken answers and return feedback against the answer you built from your own material. You can then revisit missed points as cards or listen to a topic as an examiner-voiced lecture. Start from your own notes at portal.mysummaries.app.