Exercises — Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

WHAT law? Currents at a junction → this is charge conservation, so we use KCL. Apply the sign rule (in $+$, out $-$): $I_1+I_2-I_3=0.$ Solve: $I_3=I_1+I_2=4+1=\boxed{5\text{ A}}.$ Sanity: everything that flows in must flow out — no charge is stored at a bare junction.

WHAT is happening? Current flows from high potential to low potential through a resistor (that is what "resistance" does — it burns energy). Walking with the current means walking downhill in potential. So it is a drop, written with a minus sign: $\Delta V=-IR=-(2)(3)=\boxed{-6\text{ V}}.$ Had you walked against the current, the same magnitude would appear as $+6\text{ V}$ (a rise).

Step 1 — assume a direction. Say $I$ flows clockwise. (A wrong guess only flips the sign of the answer.) Step 2 — apply KVL starting at the battery's $-$ terminal, going clockwise: $+\epsilon -I{R}_1-I{R}_2=0.$ Why these signs? $-$ to $+$ inside the battery is a rise ($+\epsilon$); each resistor traversed with the current is a drop ($-IR$). Step 3 — solve: $I=\frac{\epsilon }{R_1+R_2}=\frac{9}{3}=\boxed{3\text{ A}}.$ Cross-check with Series and Parallel Resistors: series resistance $=3\Omega$, so $I=9/3=3\text{ A}$. ✓

Using $I=3\text{ A}$ from before and Ohm's Law: $V_1=I{R}_1=3\times 1=3\text{ V},V_2=I{R}_2=3\times 2=6\text{ V}.$ Check via KVL: $V_1+V_2=3+6=9\text{ V}=\epsilon .$ ✓ This is exactly the Conservation of Energy statement: all the energy the battery gives each coulomb ($9\text{ J/C}$) is spent in the two resistors.

KVL on each branch loop — both resistors sit directly across the battery, so both see $\epsilon$: $I_1=\frac{\epsilon }{R_1}=\frac{12}{6}=2\text{ A},I_2=\frac{\epsilon }{R_2}=\frac{12}{3}=4\text{ A}.$ KCL at the top node (total splits into two branches): $I=I_1+I_2=2+4=\boxed{6\text{ A}}.$ Check: $R_{\text{eq}}=\frac{R_1{R}_2}{R_1+R_2}=\frac{6\cdot 3}{9}=2\Omega$, so $I=12/2=6\text{ A}$. ✓ Note the smaller resistor ($R_2$) carries the larger current — current prefers the easy path.

KCL (in $+$, out $-$): $I_1-I_2-I_3=0\Rightarrow I_3=I_1-I_2=5-8=\boxed{-3\text{ A}}.$ Interpret the minus sign: we assumed $I_3$ flows out, but it came out negative. That means the real current flows into the node with magnitude $3\text{ A}$. Sanity: then in-flow $=5+3=8\text{ A}$, out-flow $=8\text{ A}$. Balanced. ✓

Step 1 — collapse the parallel pair (from Series and Parallel Resistors): $R_p=\frac{R_2{R}_3}{R_2+R_3}=\frac{6\cdot 3}{9}=2\Omega .$ Step 2 — KVL on the outer loop (with $I$ = total current through $R_1$ and $R_p$): $\epsilon -I{R}_1-I{R}_p=0\Rightarrow I=\frac{\epsilon }{R_1+R_p}=\frac{14}{2+2}=\boxed{3.5\text{ A}}.$ Step 3 — voltage across the parallel pair (the "look at the mint node" in the figure): $V_p=I{R}_p=3.5\times 2=7\text{ V}.$ Step 4 — split via KCL + Ohm (both branches see $V_p=7\text{ V}$): $I_2=\frac{V_p}{R_2}=\frac{7}{6}\approx 1.1667\text{ A},I_3=\frac{V_p}{R_3}=\frac{7}{3}\approx 2.3333\text{ A}.$ KCL check at the split node: $I_2+I_3=\frac{7}{6}+\frac{7}{3}=\frac{7}{6}+\frac{14}{6}=\frac{21}{6}=3.5\text{ A}=I.$ ✓

Assume $I$ clockwise in the direction ${\epsilon }_1$ pushes. Travelling clockwise:

• through ${\epsilon }_1$ ($-$ to $+$): rise $+{\epsilon }_1$,
• through ${\epsilon }_2$ ($+$ to $-$, since it opposes): drop $-{\epsilon }_2$,
• through $R$ with the current: drop $-IR$. KVL: ${\epsilon }_1-{\epsilon }_2-IR=0\Rightarrow I=\frac{{\epsilon }_1-{\epsilon }_2}{R}=\frac{10-4}{3}=\boxed{2\text{ A}}.$ The stronger battery wins; the net driving EMF is $10-4=6\text{ V}$.

Balance condition. The bridge is balanced when the two midpoints (B and D in the figure) sit at the same potential, so no current crosses the galvanometer. That happens when $\frac{R_1}{R_2}=\frac{R_3}{R_4}.$ Check: $\frac{R_1}{R_2}=\frac{2}{4}=0.5$ and $\frac{R_3}{R_4}=\frac{3}{6}=0.5$. Equal → balanced. ✓ WHY this ratio? With the galvanometer carrying zero current (KCL forces the current through $R_1$ to equal that through $R_2$, and likewise $R_3=R_4$ pair), each side is a simple series voltage divider. Equal ratios put both midpoints at the same fraction of $\epsilon$ → equal potential → no bridge current (this uses Electric Potential single-valuedness). Total current (galvanometer open, so the two branches are independent, then in parallel): $\text{Left branch: }{R}_1+R_2=6\Omega ,\text{Right branch: }{R}_3+R_4=9\Omega .$ $R_{\text{eq}}=\frac{6\cdot 9}{6+9}=\frac{54}{15}=3.6\Omega ,I=\frac{\epsilon }{R_{\text{eq}}}=\frac{15}{3.6}=\boxed{4.1\overline{6}\text{ A}}.$

This is Mesh and Nodal Analysis — a systematic KVL application. The shared resistor $R_m$ carries the difference $(I_a-I_b)$ because the two clockwise loops push through it in opposite directions. KVL, left loop (source + $R_a$ + shared $R_m$): $\epsilon -I_a{R}_a-(I_a-I_b)R_m=0.$ Plug numbers: $20-4I_a-2(I_a-I_b)=0\Rightarrow 6I_a-2I_b=20.$ KVL, right loop ($R_b$ + shared $R_m$, no source): $-(I_b-I_a)R_m-I_b{R}_b=0\Rightarrow -2(I_b-I_a)-6I_b=0\Rightarrow 2I_a-8I_b=0\Rightarrow I_a=4I_b.$ Substitute $I_a=4I_b$ into $6I_a-2I_b=20$: $6(4I_b)-2I_b=20\Rightarrow 22I_b=20\Rightarrow I_b=\frac{10}{11}\approx 0.9091\text{ A}.$ $I_a=4I_b=\frac{40}{11}\approx 3.6364\text{ A}.$ Current in the shared resistor: $I_a-I_b=\frac{40}{11}-\frac{10}{11}=\frac{30}{11}\approx 2.7273\text{ A}$ (flowing in the left-loop's clockwise sense). ✓

One line each — the reflexes this set is training. When is KCL the right tool? ::: When charge meets or splits at a node — sum of currents $=0$. When is KVL the right tool? ::: When you can trace a closed loop — sum of signed potential changes $=0$. What does a negative current mean? ::: Assumed arrow was backward; keep the magnitude, flip the direction. Sign of a resistor traversed with the current? ::: A drop, $-IR$. Sign of a battery entered at its − terminal? ::: A rise, $+\epsilon$. Current in a resistor shared by two clockwise mesh loops? ::: The difference of the two mesh currents. Wheatstone balance condition? ::: $R_1/R_2=R_3/R_4$ → equal midpoint potentials → zero galvanometer current.