123-ME · Series/Parallel Networks & Dividers

Series/Parallel Networks & Dividers

Slides 2.1–2.3 · reduce, solve, back-substitute

Real circuits are rarely a single loop. They're tangles of resistors that look intimidating until you notice most of that tangle is just series and parallel combinations wearing a disguise. This chapter gives you the two reduction rules ( $R_{eq}$ for series, $R_{eq}$ for parallel), the five-step procedure for chaining them across an entire network, and the two shortcuts — voltage and current dividers — that let you skip the reduction step entirely once you recognize the pattern. By the end you should be able to look at a network, know in what order to collapse it, and read off any voltage or current without solving a single simultaneous equation. Budget about 20 minutes here; the payoff is that Chapter 2.4 onward (node-voltage, mesh-current) becomes the fallback for circuits that can't be reduced this way — you'll want this chapter's tools reflexively so you reach for the harder machinery only when you actually need it.

1 Series resistors: same current, voltages add

Two elements are in series when they share a single path — no other element connects to the node between them, so whatever current enters one must be the current that leaves it and enters the next. That's the defining property, not "drawn in a line":

Series current constraint

i_a = i_b = i_c = i

Now derive $R_{eq}$ instead of just stating it. Walk KVL around the loop, summing drops in the direction of current flow, then apply Ohm's law to each element individually:

Series R_eq derivation

V_s = V_1 + V_2 + V_3 = iR_1 + iR_2 + iR_3 = i(R_1+R_2+R_3)

\text{Compare to } V_s = iR_{eq} \quad\Rightarrow\quad R_{eq} = R_1 + R_2 + R_3

Because $i$ is the same in every term, it factors straight out — that's the whole reason series resistances simply add. It generalizes to any count: $R_{eq} = \sum_n R_n$ . The equivalent resistor is a stand-in that draws the exact same current from the source as the whole chain did; nothing else about the circuit needs to know the chain has been collapsed.

2 Parallel resistors: same voltage, currents add

Two elements are in parallel when they share both endpoints — the same pair of nodes — so by KVL around the loop formed by either two of them, they must have the identical voltage across them:

Parallel voltage constraint

v_1 = v_2 = v_3 = v

Derive the parallel rule from KCL at the top node — total current in equals the sum of branch currents out:

Parallel R_eq derivation

i = i_1+i_2+i_3 = \frac{v}{R_1}+\frac{v}{R_2}+\frac{v}{R_3} = v\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)

\text{Compare to } i = \frac{v}{R_{eq}} \quad\Rightarrow\quad \frac{1}{R_{eq}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

This time $v$ is what's shared, so the conductances ( $G = 1/R$ ) add instead: $G_{eq} = G_1+G_2+G_3$ . For exactly two resistors this collapses to the handy product-over-sum shortcut:

R_{eq} = \frac{R_1 R_2}{R_1+R_2} \qquad \text{(two resistors only — don't use this form for three or more)}

Sanity checks worth memorizing

Parallel $R_{eq}$ is always smaller than the smallest branch resistance — adding a path can only make it easier for current to flow. Series $R_{eq}$ is always larger than the biggest single resistor. If a calculation gives you the opposite, you've mixed up the two rules.

3 Combining series and parallel: the reduction procedure

Most real networks are neither purely series nor purely parallel — they're a nested mix. The slides' five-step procedure handles any such network, provided it genuinely reduces to series/parallel groups (once you hit a bridge or a network with no such reduction, that's your cue for node-voltage or mesh-current analysis in the next chapter):

The five-step procedure

Locate a series or parallel combination — usually easiest starting farthest from the source.
Redraw the circuit with that group replaced by its single equivalent resistor.
Repeat steps 1–2 until the whole network is one $R_{eq}$ in series with the source.
Solve the simplified (single-loop) circuit for the source current, then work backward substituting into each un-collapsed stage to recover every original voltage/current.
Verify: KCL at each node, KVL around each loop, and total power delivered = total power absorbed.

Worked Example 3.1 — nested reduction, 90 V source

A 90 V source feeds $R_1 = 10\ \Omega$ in series, then $R_2=30\ \Omega \parallel R_3=60\ \Omega$ . Find the source current and every branch current, then check power balance.

Reduce the parallel pair first (farthest from source):

R_{eq1} = \frac{30\times60}{30+60} = 20\ \Omega

Now it's a single loop: $R_{eq} = R_1 + R_{eq1} = 10+20 = 30\ \Omega$ .

i_1 = \frac{90}{30} = 3\text{ A (source/}R_1\text{ current)}

Back-substitute: the voltage across the parallel pair is $V_2 = i_1 R_{eq1} = 3\times20 = 60$ V, the same voltage seen by $R_2$ and $R_3$ individually.

i_2 = \frac{60}{30}=2\text{ A}, \qquad i_3=\frac{60}{60}=1\text{ A} \qquad (\text{check: } i_2+i_3 = 3\text{ A} = i_1\ \checkmark)

Power balance: source delivers $P_s = 90\times3 = 270$ W. Resistors absorb $i_1^2R_1 + i_2^2R_2 + i_3^2R_3 = 9(10)+4(30)+1(60) = 90+120+60=270$ W. Matches. ✓

4 Voltage divider: splitting a voltage without solving anything

Once you recognize a pure series chain, you don't even need to grind through the full procedure to find the voltage across one element — it falls straight out of the derivation above. Since every resistor in the chain carries the same current $i = V_{\text{total}}/R_{eq}$ , the voltage across any one of them is:

Voltage divider derivation

i = \frac{V_{\text{total}}}{R_1+R_2+\cdots+R_N} \quad\Rightarrow\quad V_n = iR_n = \frac{R_n}{R_1+R_2+\cdots+R_N}\,V_{\text{total}}

Read this as a proportion: each resistor claims a share of the total voltage exactly proportional to its own resistance relative to the whole chain. The biggest resistor in the chain drops the most voltage — it's the one "fighting" the current the hardest.

Worked Example 3.2 — Arduino sensor divider (real-world application)

A car's electrical system nominally sits at 12 V but can fluctuate up to about 14.8 V while the alternator is charging. An Arduino analog input can only tolerate up to 5 V — feeding it the raw battery voltage would destroy the pin. A voltage divider scales the signal down safely before it ever reaches the microcontroller.

Pick $R_1 = 2\ \text{k}\Omega$ (top, connected to $V_{\text{batt}}$ ) and $R_2 = 1\ \text{k}\Omega$ (bottom, to ground) and tap $V_{out}$ across $R_2$ :

V_{out} = \frac{R_2}{R_1+R_2}V_{\text{batt}} = \frac{1}{3}V_{\text{batt}}

Check both ends of the operating range against the Arduino's 5 V ceiling:

V_{out}(12\text{ V}) = \frac{1}{3}(12)=4.0\text{ V}, \qquad V_{out}(14.8\text{ V}) = \frac{1}{3}(14.8)\approx4.93\text{ V}

Both comfortably under 5 V — the pin is safe across the whole fluctuation range.

Check the design doesn't waste much power: with $R_1+R_2 = 3\ \text{k}\Omega$ ,

i = \frac{12}{3000} = 4\text{ mA}, \qquad P = i^2(R_1+R_2) = (0.004)^2(3000) = 48\text{ mW}

A trickle — fine for a sensor tap that's always connected.

V_{out}=\tfrac{1}{3}V_{\text{batt}} \;\Rightarrow\; 4.0\text{–}4.93\text{ V, safely below the 5 V limit, at 48 mW}

5 Current divider: splitting a current between parallel branches

The mirror-image idea for parallel branches: since every branch sees the same voltage $v = i_{\text{total}}R_{eq}$ , the current through any one branch is:

Current divider derivation (two branches)

v = i_{\text{total}}R_{eq} = i_{\text{total}}\frac{R_1R_2}{R_1+R_2}

i_1 = \frac{v}{R_1} = \frac{R_2}{R_1+R_2}\,i_{\text{total}}, \qquad i_2 = \frac{v}{R_2} = \frac{R_1}{R_1+R_2}\,i_{\text{total}}

The classic mix-up

Notice branch $1$ 's current formula has $R_2$ (the other resistor) on top — current division is inversely proportional to resistance, the opposite of voltage division. The smaller resistance always grabs the larger share of the current — it's the path of least resistance, literally. Mixing this up with the voltage-divider formula is the single most common algebra slip in this topic.

Worked Example 3.3 — combined divider application

100 V source, $R_1=60\ \Omega$ in series, then $R_2=30\ \Omega \parallel R_3=60\ \Omega$ . Find the voltage across the parallel pair and the current through $R_3$ — using dividers directly, without fully solving the circuit.

Reduce the parallel pair to get the divider ratio: $R_x = \dfrac{30\times60}{30+60}=20\ \Omega$ .

Voltage divider across the now-two-element series chain ( $R_1$ and $R_x$ ):

V_x = \frac{R_x}{R_1+R_x}\times100 = \frac{20}{60+20}\times100 = 25\text{ V}

Current divider inside the parallel pair — need the total current entering it first, $i_s = V_x/R_x = 25/20 = 1.25$ A, then split by the inverse-resistance rule:

i_3 = \frac{R_2}{R_2+R_3}\,i_s = \frac{30}{30+60}\times1.25 \approx 0.417\text{ A}

V_x = 25\text{ V}, \qquad i_3 \approx 0.417\text{ A}

6 Interactive: build your own network

Click a branch to select it, then use the buttons to add a resistor in series with it or split it into a parallel pair. Watch $R_{eq}$ recompute live as the topology changes — this is exactly the reduction procedure from Section 3, just automated so you can build intuition fast.

Clickable Resistor Network

Selected branch

branch 1 (10 Ω)

Equivalent resistance seen by the source: R_eq = 10.0 Ω

7 Self-check

Self-check

A 10 V source feeds a 6 Ω series resistor, then two 8 Ω resistors in parallel. What is the current through each 8 Ω resistor?

Self-check

Three resistors 5 Ω, 7 Ω, 13 Ω are in series across a 25 V source. Which resistor has the largest voltage across it, and why?

8 Practice problems

Problem 3.1

series-parallel reduction

A 48 V source feeds $R_1=4\ \Omega$ in series with the parallel combination of $R_2=12\ \Omega$ , $R_3=6\ \Omega$ , and $R_4=4\ \Omega$ . Find the source current $i_s$ and the current through $R_3$ .

Method 1 — direct reduction. Combine the parallel trio via conductances: $G_{eq}=\tfrac{1}{12}+\tfrac{1}{6}+\tfrac{1}{4} = \tfrac{1+2+3}{12}=\tfrac{6}{12}=0.5\ \text{S} \Rightarrow R_{eq1}=2\ \Omega$ . Total $R_{eq}=R_1+R_{eq1}=4+2=6\ \Omega$ . Source current $i_s = 48/6 = 8$ A. Voltage across the parallel group: $V=i_sR_{eq1}=8\times2=16$ V. Current through $R_3$ : $i_3 = 16/6 \approx 2.667$ A.

Method 2 — current-divider cross-check. Skip computing $V$ explicitly: use the current-divider formula directly on the three-branch parallel group. For branch $R_3$ among parallel branches $R_2,R_3,R_4$ , the share is proportional to that branch's conductance over the total: $i_3 = \dfrac{G_3}{G_2+G_3+G_4}\,i_s = \dfrac{1/6}{1/12+1/6+1/4}\times8 = \dfrac{1/6}{1/2}\times8 = \dfrac{1}{3}\times8 \approx 2.667$ A. Matches Method 1. ✓

i_s = 8\text{ A}, \qquad i_3 \approx 2.667\text{ A}

Problem 3.2

voltage divider

A 36 V source drives four series resistors: $R_1=2\ \Omega$ , $R_2=4\ \Omega$ , $R_3=6\ \Omega$ , $R_4=8\ \Omega$ . Find $V_3$ , the voltage across $R_3$ , two independent ways.

Method 1 — voltage-divider formula. $R_{total}=2+4+6+8=20\ \Omega$ . $V_3 = \frac{R_3}{R_{total}}V_s = \frac{6}{20}\times36 = 10.8\text{ V}$

Method 2 — direct current calculation (independent check). Find the loop current first: $i = V_s/R_{total} = 36/20 = 1.8$ A. Then Ohm's law on $R_3$ alone: $V_3 = iR_3 = 1.8\times6 = 10.8$ V. Matches Method 1. ✓

V_3 = 10.8\text{ V}

Problem 3.3

current divider, three branches

A 9 A current source feeds three parallel branches: $R_1=6\ \Omega$ , $R_2=3\ \Omega$ , $R_3=9\ \Omega$ . Find the current through $R_2$ two independent ways.

Method 1 — reduce to voltage, then Ohm's law. $G_{eq} = \tfrac16+\tfrac13+\tfrac19 = \tfrac{3+6+2}{18}=\tfrac{11}{18}\ \text{S} \Rightarrow R_{eq}=\tfrac{18}{11}\ \Omega$ . Shared voltage: $v = i_sR_{eq} = 9\times\tfrac{18}{11} = \tfrac{162}{11}\approx14.727$ V. Then $i_2 = v/R_2 = 14.727/3 \approx 4.909$ A.

Method 2 — general current-divider formula (independent check). For branch $k$ among parallel branches, $i_k = \dfrac{G_k}{\sum G}\,i_s$ . Here $i_2 = \dfrac{1/3}{11/18}\times9 = \dfrac{6/18}{11/18}\times9 = \dfrac{6}{11}\times9 = \dfrac{54}{11}\approx4.909$ A. Matches Method 1. ✓

i_2 \approx 4.909\text{ A}