Dawson 203-SN2-RE tutoring

1. Define a sign convention. Let $U$ be the potential at junction $J$, relative to the reference. Define $I_1$ from the first source toward $J$, $I_2$ from the second source toward $J$, and $I_3$ from $J$ toward the reference. Ohm’s law uses the potential at the start of each arrow minus the potential at its end: $I_1=(12\,\mathrm{V}-U)/(6\,\Omega)$, $I_2=(6\,\mathrm{V}-U)/(3\,\Omega)$, and $I_3=U/(3\,\Omega)$. 2. Apply current conservation at the junction. Since charge does not accumulate at $J$, $I_1+I_2=I_3$. Hence $\frac{12\,\mathrm{V}-U}{6\,\Omega}+\frac{6\,\mathrm{V}-U}{3\,\Omega}=\frac{U}{3\,\Omega}$. Multiplying by $6\,\Omega$ gives $12\,\mathrm{V}-U+2(6\,\mathrm{V}-U)=2U$, or $24\,\mathrm{V}=5U$. Thus $U=4.8\,\mathrm{V}$. 3. Compute and check the currents. Substitution gives $I_1=1.2\,\mathrm{A}$, $I_2=0.4\,\mathrm{A}$, and $I_3=1.6\,\mathrm{A}$. The check $1.2+0.4=1.6$ confirms the junction balance. The voltage across each source-branch resistor is the difference between the potentials at its two ends. It is not the source’s full potential relative to the reference. 4. Recalculate after changing the source. Use the same arrows and replace only the second source potential. The balance becomes $\frac{12\,\mathrm{V}-U}{6\,\Omega}+\frac{3\,\mathrm{V}-U}{3\,\Omega}=\frac{U}{3\,\Omega}$, giving $18\,\mathrm{V}=5U$, so $U=3.6\,\mathrm{V}$. Now $I_2=(3-3.6)\,\mathrm{V}/(3\,\Omega)=-0.2\,\mathrm{A}$. Its actual direction is opposite the chosen arrow: a current of $0.2\,\mathrm{A}$ flows from $J$ toward the second source. The other currents are $I_1=1.4\,\mathrm{A}$ and $I_3=1.2\,\mathrm{A}$, and $1.4-0.2=1.2$ still balances. A negative signed current is a direction result, not automatically an algebra error.