Exercises — Series vs parallel resistor combinations

WHAT: end-to-end, one path ⇒ same current flows through both ⇒ series. WHY add: with the same current $I$, KVL says the drops stack: $V=I{R}_1+I{R}_2=I(R_1+R_2)$. $R_{eq}=4+6=10\Omega$ Sanity: series result ($10$) is bigger than the largest resistor ($6$). ✓

WHAT: both span the same two nodes ⇒ same voltage across each ⇒ parallel. WHY product-over-sum: conductances add, and for two resistors that algebra collapses to $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{4\times 6}{4+6}=\frac{24}{10}=2.4\Omega$ Sanity: parallel result ($2.4$) is smaller than the smallest resistor ($4$). ✓

One resistor is its own equivalent: $R_{eq}=8\Omega$. No law needed — there is nothing to combine. This is the degenerate case of both formulas with $n=1$.

Same current through all three ⇒ add: $R_{eq}=100+220+330=650\Omega$ Bigger than the largest ($330$). ✓

WHY the shortcut $R/n$: each identical path carries an equal share of current, so $n$ paths let $n\times$ the current flow at the same voltage — resistance divides by $n$. $\frac{1}{R_{eq}}=\frac{1}{90}+\frac{1}{90}+\frac{1}{90}=\frac{3}{90}\Rightarrow R_{eq}=\frac{90}{3}=30\Omega$ Smaller than $90$. ✓

Combine first: $R_{eq}=100+200=300\Omega$. WHY $R_{eq}$ first: the source only "sees" the equivalent, so total current is $I=V/R_{eq}$. $I=\frac{12}{300}=0.04\text{A}$ Same current flows through $R_2$, so by Ohm's Law (this is a Voltage Divider): $V_2=I{R}_2=0.04\times 200=8\text{V}$ Check: $V_1=0.04\times 100=4\text{V}$, and $4+8=12$ ✓ (KVL).

Same $12\text{V}$ across each (that's the definition of parallel): $I_1=\frac{12}{100}=0.12\text{A},I_2=\frac{12}{200}=0.06\text{A}$ By KCL the source current is the sum (Current Divider in action): $I=I_1+I_2=0.18\text{A}$ Cross-check via $R_{eq}=\frac{100\times 200}{300}=66.\overline{6}\Omega$: $I=12/66.\overline{6}=0.18\text{A}$ ✓

Step A — collapse the innermost parallel pair (the branch node, see figure): $R_{23}=\frac{30\times 60}{30+60}=\frac{1800}{90}=20\Omega$ Step B — now it's a simple series chain $R_1$ then $R_{23}$: $R_{eq}=10+20=30\Omega$ WHY this order: like nested parentheses, evaluate the inside first. $R_{23}$ ($20$) is below the smallest of its pair ($30$) ✓; then the series step pushes it up to $30$.

Total current from the source, using $R_{eq}=30\Omega$: $I=\frac{9}{30}=0.3\text{A}$ This whole current passes through $R_1$ (series), so its drop is $V_1=0.3\times 10=3\text{V}$. By KVL the parallel block gets the rest: $V_{23}=9-3=6\text{V}$ (also $=I\times R_{23}=0.3\times 20$ ✓). That $6\text{V}$ sits across $R_3$: $I_3=\frac{6}{60}=0.1\text{A}$ Check: $I_2=6/30=0.2\text{A}$, and $I_2+I_3=0.3=I$ ✓ (KCL).

Step A — collapse each branch to one number. Branch A is series: $5+5=10\Omega$. Branch B is already $20\Omega$. Step B — now two resistors in parallel ($10\Omega \parallel 20\Omega$): $R_{eq}=\frac{10\times 20}{10+20}=\frac{200}{30}=6.\overline{6}\Omega$ Below the smaller branch ($10$) ✓.

Set the parallel formula equal to the target and solve for $R$: $\frac{1}{8}=\frac{1}{12}+\frac{1}{R}\Rightarrow \frac{1}{R}=\frac{1}{8}-\frac{1}{12}=\frac{3-2}{24}=\frac{1}{24}$ $R=24\Omega$ Sanity: $8<12$, so parallel (which only lowers resistance) is the right choice. Check: $\frac{12\times 24}{36}=\frac{288}{36}=8\Omega$ ✓.

Series adds, so the extra resistor is just the shortfall: $R=8-5=3\Omega$ WHY series here: the target ($8$) is larger than the resistor you have ($5$), and only series can raise resistance. If the target were below $5$, series would be impossible and you'd need parallel instead.

Idea: two in parallel give $12/2=6\Omega$; put that in series with the third $12\Omega$. $R_{eq}=\underset{\text{two in parallel}}{\underbrace{\frac{12}{2}}}+12=6+12=18\Omega$ WHY it works: the parallel pair drops below $12$ (to $6$), then the series resistor lifts it back up past $12$ to land on $18$. Mixing "down then up" is how you hit in-between targets.

Divide top and bottom by $R_2$: $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{R_1}{\frac{R_1}{R_2}+1}\underset{R_2\to \infty }{\to }\frac{R_1}{0+1}=R_1$ Physical meaning: an infinite resistor is a broken path — no current takes it, so the network behaves as if only $R_1$ is present. A parallel branch that carries no current is invisible. ✓

$R_{eq}=\frac{R_1\cdot 0}{R_1+0}=\frac{0}{R_1}=0\Omega$ Physical meaning: a zero-resistance wire in parallel is a short circuit — it offers a free path, so all the current rushes through it and none through $R_1$. Two nodes bridged by a wire become one node; the resistance between them is $0$. This is the extreme of "parallel drops below the smallest."

Balance argument: by symmetry, the two midpoints sit at the same voltage (each side is an identical $R$–$R$ Voltage Divider splitting the input in half). Equal voltage across $R_5$ means zero voltage difference, so by Ohm's Law $I_5=V/R_5=0/R_5=0$: no current flows through the bridge. Consequence: since $R_5$ carries no current, we can delete it (or short it) without changing anything. The network becomes two series pairs in parallel:

• top path: $R_1+R_2=R+R=2R$
• bottom path: $R_3+R_4=R+R=2R$ $R_{eq}=\frac{(2R)(2R)}{2R+2R}=\frac{4R^2}{4R}=R$ WHY this is beautiful: the messy 5-resistor network collapses to a single $R$ purely from a symmetry (equal-voltage) argument — no simultaneous equations needed. For $R=10\Omega$: $R_{eq}=10\Omega$.

Series shares the same current; parallel shares the same voltage. Series $R_{eq}$ is always larger than any resistor; parallel $R_{eq}$ is always smaller than the smallest. An infinite parallel branch acts like an open circuit (invisible); a zero-ohm parallel branch acts like a short ($R_{eq}=0$). In a balanced bridge the middle resistor carries zero current.