Study guides

Electrical Circuits

5 guides for this subject — how to prepare, what to practise, and how to know it has gone in.

A practice question

Choose an answer and it is marked, with the explanation. From Electrical Circuits Practice Questions: 8 Worked MCQs.

Question 11 mark

A 12 V ideal source is connected in series with 2 kΩ and 4 kΩ resistors. What is the current through the 4 kΩ resistor?

The series resistance is 2 kΩ + 4 kΩ = 6 kΩ, so I = V/R = 12/6000 = 0.002 A = 2 mA. The strongest distractor is 3 mA, which comes from incorrectly using only the 4 kΩ resistor and ignoring the other series resistor.

A spoken answer, marked

An examiner's question, answered out loud and marked criterion by criterion. From Electrical Circuits Viva Questions: Three Worked Practice Stations.

Oral — A loaded resistive dividerMarked

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.

2:363:00Mark answer
76%A loaded resistive divider — marked76/100 · Developing · 2:36 spoken of 3:00
Circuit model and assumptions15/20

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.

Equation and calculation16/20

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.

Units and interpretation15/20

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.

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