McGill PHYS 102: Introductory Physics — Electromagnetism tutoring

A node is a connected region of ideal wire at one potential. Trace the two endpoints of each resistor before choosing a series or parallel rule. The branch currents add where the branches rejoin.

1. Read the connections. $R_2$ and $R_3$ share both endpoints, so they have the same voltage and are in parallel. $R_1$ carries all the source current before it splits. $R_1$ and $R_2$ do not carry the same current simply because they appear beside one another in a drawing.

2. Replace the parallel pair temporarily. Its resistance $R_p$ satisfies $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$ in $\Omega^{-1}$, so $R_p=2\,\Omega$. The total resistance is $2+2=4\,\Omega$. Therefore $I=\frac{12}{4}=3\,\mathrm{A}$ through the source and $R_1$.

3. Recover the actual branch voltage. The drop across $R_1$ is $IR_1=3\times2=6\,\mathrm{V}$, leaving $12-6=6\,\mathrm{V}$ across each parallel resistor. The 12 V source is not connected directly across either branch resistor.

4. Find and check the currents. $I_2=\frac{6}{6}=1\,\mathrm{A}$ and $I_3=\frac{6}{3}=2\,\mathrm{A}$. The junction check gives $1+2=3\,\mathrm{A}$. Around either loop, the 6 V drop across $R_1$ and the 6 V branch drop add to the source’s 12 V rise.

5. Check energy per second. The source supplies $12\times3=36\,\mathrm{W}$. The resistors dissipate $3^2\times2=18\,\mathrm{W}$, $1^2\times6=6\,\mathrm{W}$ and $2^2\times3=12\,\mathrm{W}$. Their sum is 36 W. This second check can catch a branch current that passed an isolated arithmetic check.

Common mistake: assigning 12 V to every resistor produces branch currents for a different circuit. Identify the nodes before inserting a voltage into $I=\frac{V}{R}$.