Université de Montréal PHY1902 tutoring

1. Identify the connections. The parallel resistors connect to the same two points, called nodes, so the voltage across them is the same. Let $R_p$ be their equivalent resistance. Then $\frac{1}{R_p}=\frac{1}{6\,\Omega}+\frac{1}{3\,\Omega}=\frac{1}{2\,\Omega}$, hence $R_p=2\,\Omega$. The series resistor adds directly, giving total resistance $R_{\mathrm{tot}}=4\,\Omega+2\,\Omega=6\,\Omega$. 2. Find the source current. Ohm’s law states $I=U/R$, where $I$ is current, $U$ the voltage across a resistance, and $R$ that resistance. Thus the source current is $I_s=(12\,\mathrm{V})/(6\,\Omega)=2\,\mathrm{A}$. It also flows through the series resistor, whose voltage drop is $U_4=I_s(4\,\Omega)=8\,\mathrm{V}$. 3. Find the branch currents. The remaining voltage across the parallel pair is $U_p=12\,\mathrm{V}-8\,\mathrm{V}=4\,\mathrm{V}$. Let $I_6$ and $I_3$ be the currents through the resistors of $6\,\Omega$ and $3\,\Omega$, respectively. Then $I_6=(4\,\mathrm{V})/(6\,\Omega)=\frac{2}{3}\,\mathrm{A}$ and $I_3=(4\,\mathrm{V})/(3\,\Omega)=\frac{4}{3}\,\mathrm{A}$. 4. Check that the branch currents add up to the source current at the junction: $I_6+I_3=2\,\mathrm{A}=I_s$. The loop voltage check is $8\,\mathrm{V}+4\,\mathrm{V}=12\,\mathrm{V}$. Applying the full source voltage to each parallel branch would fail these checks because it would ignore the series voltage drop. The smaller branch resistance carries twice the current at the same voltage. Doubling the source voltage would double every current in this ideal resistive model.