John Abbott College · 203-SN2-RE

John Abbott 203-SN2-RE: Electricity and Magnetism Tutoring

A circuit diagram can look straightforward until the current splits or the voltage changes between components. Nablio helps you turn that diagram into equations, track signs and units, and explain why a result makes physical sense. Bring a problem from your current 203-SN2-RE work so your session can focus on the step that is blocking you.

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What we can work on

Electric charge and fields: separate the field created by a source from the force on a test charge.
Resistors and circuits: identify series and parallel connections, then check currents and voltage drops.
Capacitors: connect charge, capacitance and voltage without confusing capacitor rules with resistor rules.
Magnetic fields and forces: draw directions before calculating magnitudes.

A useful starting point for circuit questions

Mark nodes before simplifying the circuit. Two resistors are in parallel only if they connect to the same two nodes; components in series must carry the same current without a branch between them. Once you have a result, check that current entering a junction equals current leaving it and that the voltage drops around a loop balance the source.

Worked example: a series resistor feeding two branches

An ideal 12 V source is connected to a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors. Find the source current and the current in each parallel branch. Assume ideal wires.

Combine the parallel pair: $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}$ , so $R_p=2\,\Omega$ .
Add the series resistor: $R_{\mathrm{total}}=4+2=6\,\Omega$ . The source current is $I=12/6=2\,\mathrm{A}$ .
The 4 Ω resistor drops $V=IR=2\times4=8\,\mathrm{V}$ . The parallel pair therefore has 12 − 8 = 4 V across it.
Each branch has that same 4 V: $I_6=\frac{4}{6}=\frac{2}{3}\,\mathrm{A}$ and $I_3=\frac{4}{3}\,\mathrm{A}$ .
Check the junction: $\frac{2}{3}+\frac{4}{3}=2\,\mathrm{A}$ , matching the source current.

Parallel branches share a voltage, not necessarily a current. The lower-resistance branch carries more current.

Try it yourself

Keep the 12 V source and the series 4 Ω resistor, but replace the parallel branches with 12 Ω and 6 Ω. What are the source current, branch voltage and two branch currents?

Show the explained answer

The parallel resistance is 4 Ω, so the total is 8 Ω. The source current is 1.5 A. The parallel pair has 6 V across it, giving 0.5 A in the 12 Ω branch and 1.0 A in the 6 Ω branch. Their sum is 1.5 A.

Original Nablio example and practice, independent of the college. These are not course assessment questions. Follow your current section’s outline for the material to review.

Bring the diagram and your attempt

Send the question, your circuit sketch and the equations you have tried. We can work through the reasoning, then change one part of the problem to check that you can use the method independently.

We use permitted practice without completing graded assignments or exams on your behalf.

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