Exercises — Series and parallel resistance

WHAT to look for: one single path, no junction where current can split → this is series. WHY: the same current must pass through $R_1$ then $R_2$ (charge cannot pile up, KCL). Compute: resistances add, $R_{eq}=10+15=25\Omega$. Sanity: $25\Omega$ is bigger than either resistor — correct for series.

WHAT: both ends tied to the same two nodes → current can split into two branches → parallel. WHY: each resistor feels the identical node-to-node voltage. Compute (product over sum): $R_{eq}=\frac{10\times 15}{10+15}=\frac{150}{25}=6\Omega$. Sanity: $6\Omega$ is smaller than the smallest ($10\Omega$) — correct for parallel.

Top diagram: a single line, current has nowhere else to go → series. Bottom diagram: at node A the current splits into two roads and re-joins at node B → parallel. The trick is never the drawing's shape — it is whether a junction splits the current.

(a) Series → add: $R_{eq}=4+8=12\Omega$. (b) Ohm on the whole chain: $I=\frac{V}{R_{eq}}=\frac{12}{12}=1\text{A}$. WHY: the equivalent resistor draws the same current the battery really delivers. (c) Each drop is $I{R}_k$: $V_1=1\times 4=4\text{V}$, $V_2=1\times 8=8\text{V}$. KVL check: $4+8=12\text{V}$ ✓ — the drops add to the applied voltage.

(a) Product over sum: $R_{eq}=\frac{4\times 8}{4+8}=\frac{32}{12}=\frac{8}{3}\approx 2.67\Omega$. (b) Each branch feels the full $12\text{V}$ (same nodes): $I_1=\frac{12}{4}=3\text{A}$, $I_2=\frac{12}{8}=1.5\text{A}$. (c) KCL — branches add: $I=3+1.5=4.5\text{A}$. Cross-check: $I=\frac{V}{R_{eq}}=\frac{12}{8/3}=4.5\text{A}$ ✓. Note the smaller resistor ($4\Omega$) grabs the larger current.

(a) $n$ equal in series → $nR=5\times 20=100\Omega$. (b) $n$ equal in parallel → $R/n=20/5=4\Omega$. WHY: five equal roads share the load, cutting resistance to a fifth.

Step A — collapse the parallel pair. Two equal $10\Omega$ → $R/n=10/2=5\Omega$. WHY: they share the same two nodes. Step B — series with $R_1$. The whole current from the battery passes through $R_1$ then the collapsed block: $R_{eq}=5+5=10\Omega$. Battery current: $I=\frac{15}{10}=1.5\text{A}$. Voltage across the parallel block: $V_{block}=I\times 5=1.5\times 5=7.5\text{V}$ (the other $7.5\text{V}$ drops on $R_1$; KVL: $7.5+7.5=15$ ✓). Branch currents: each $=\frac{7.5}{10}=0.75\text{A}$. Check: $0.75+0.75=1.5\text{A}$ ✓.

Step A: parallel of $6$ and $3$: $\frac{6\times 3}{6+3}=\frac{18}{9}=2\Omega$. Step B: series: $R_{eq}=2+8=10\Omega$. Current: $I=\frac{20}{10}=2\text{A}$. Voltage across parallel block: $V=I\times 2=2\times 2=4\text{V}$. (Drop on $R_3$ is $2\times 8=16\text{V}$; $4+16=20$ ✓.)

Step A — inner series: $R_3+R_4=1+3=4\Omega$ (same current flows through both). Step B — parallel: that $4\Omega$ branch parallel with $R_2=6\Omega$: $\frac{4\times 6}{4+6}=\frac{24}{10}=2.4\Omega$. Step C — outer series: $R_{eq}=2+2.4=4.4\Omega$. Current: $I=\frac{12}{4.4}=\frac{30}{11}\approx 2.73\text{A}$.

Goal-driven reasoning: $9\Omega$ is between $6\Omega$ (one) and $18\Omega$ (three in series), so we need something that adds up. Put two in parallel: $6/2=3\Omega$. Then put that in series with the third $6\Omega$: $3+6=9\Omega$. ✓ Wiring: third resistor in a line, feeding a node that splits into two parallel $6\Omega$ resistors.

Set up the parallel equation: $\frac{1}{4}=\frac{1}{R}+\frac{1}{12}$. Isolate: $\frac{1}{R}=\frac{1}{4}-\frac{1}{12}=\frac{3}{12}-\frac{1}{12}=\frac{2}{12}=\frac{1}{6}$. Flip: $R=6\Omega$. Check (product over sum): $\frac{6\times 12}{6+12}=\frac{72}{18}=4\Omega$ ✓.

Translate the words: $R_1+R_2=18$ and $\frac{R_1{R}_2}{R_1+R_2}=4$. From the second: $R_1{R}_2=4\times 18=72$. So we need two numbers with sum 18 and product 72. Solve $x^2-18x+72=0$ → $x=\frac{18\pm \sqrt{324-288}}{2}=\frac{18\pm 6}{2}$. $R_1=12\Omega ,R_2=6\Omega$. Check: series $12+6=18$ ✓; parallel $\frac{72}{18}=4$ ✓.

Parallel formula: $\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{0}$. The term $\frac{1}{0}$ is infinite conductance. What that means: $\frac{1}{R_{eq}}\to \infty$, so $R_{eq}\to 0\Omega$. Physically: the $0\Omega$ wire is a free road; essentially all current takes it and the $10\Omega$ resistor is bypassed (carries $\approx 0$ current). A short across any resistor kills its effect.

Parallel formula: $\frac{1}{R_{eq}}=\frac{1}{10}+\frac{1}{\infty }=\frac{1}{10}+0$. So $R_{eq}=10\Omega$ — the broken branch is as if it were not there. WHY: infinite resistance means zero current can flow that way, so it contributes nothing. Contrast with series: a break in a series branch has infinite $R_{eq}$ (the single path is cut, current stops entirely). Same broken component, opposite consequence — location decides everything.

Use $R_{eq}=\frac{10R}{10+R}$ and check every regime:

• $R\to 0$ (short): $R_{eq}\to \frac{0}{10}=0\Omega$. Matches 5.1.
• $R=10\Omega$ (equal): $R_{eq}=\frac{100}{20}=5\Omega =R_0/2$. Two equal resistors → half.
• $R\to \infty$ (open): $R_{eq}\to \frac{10R}{R}=10\Omega =R_0$. Matches 5.2. Conclusion: $R_{eq}$ rises monotonically from $0$ up toward $10\Omega$ but never reaches it. So a parallel combination is always strictly less than the smaller resistor — the boundary $10\Omega$ is a ceiling approached only as the other branch vanishes.

For any network: (1) find a pure series or pure parallel block, (2) collapse it, (3) redraw, (4) repeat until one resistor remains, then use Ohm's Law to march current and voltage back out. Series sums resistance and splits voltage; parallel sums conductance and splits current. That's the entire game.