Compute equivalent resistance in mixed networks: Exercises

Intuition How to use this page

Each problem stands alone — read it, try it on paper, then open the collapsible solution. The difficulty climbs in five steps: recognising a pattern, applying one rule, chaining several, tearing apart something disguised, and finally a full-mastery build. Every symbol we use is defined on the parent note; if you want the same content in Hindi-English, see 1.2.02 Compute equivalent resistance in mixed networks (Hinglish).

The only tools you need:

Series (share one node, nothing else attached → same current → resistances add), justified by Kirchhoff's Voltage Law.
Parallel (share both nodes → same voltage → conductances add), justified by Kirchhoff's Current Law.
Both rest on Ohm's Law: a two-terminal box always behaves like one $R_{e q}$ with $V = I R_{e q}$ .

Level 1 — Recognition

Goal: name the pattern and apply exactly one rule.

Exercise L1.1

Two resistors $R_{1} = 3 Ω$ and $R_{2} = 5 Ω$ are connected end-to-end between terminals A and B, with nothing tapping the joint between them. Find $R_{A B}$ .

Recall Solution L1.1

WHAT: the shared middle node touches only these two resistors, so the same current flows through both → series. WHY: by Kirchhoff's Voltage Law the drops add: $V = I R_{1} + I R_{2} = I (R_{1} + R_{2})$ . $R_{A B} = R_{1} + R_{2} = 3 + 5 = 8 Ω$ Series always makes resistance bigger than either piece — $8 > 5$ . ✓

Exercise L1.2

Two resistors $R_{1} = 3 Ω$ and $R_{2} = 6 Ω$ are connected side-by-side: both left ends meet at node A, both right ends meet at node B. Find $R_{A B}$ .

Recall Solution L1.2

WHAT: both resistors share both end nodes → same voltage → parallel. WHY: by Kirchhoff's Current Law the currents add: $I = \frac{V}{R _{1}} + \frac{V}{R _{2}}$ , so conductances add. For exactly two resistors use product-over-sum: $R_{A B} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2 Ω$ Parallel must land below the smallest branch — $2 < 3$ . ✓

Exercise L1.3

Three equal resistors, each $R = 12 Ω$ , are all in parallel between A and B. Find $R_{A B}$ .

Recall Solution L1.3

WHAT: three resistors, so product-over-sum is not allowed — go back to the reciprocal sum. $\frac{1}{R _{A B}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$ Flip it (the classic missed step): $R_{A B} = 4 Ω$ Shortcut worth remembering: $n$ equal resistors in parallel give $R / n = 12/3 = 4 Ω$ . ✓

Level 2 — Application

Goal: two rules, in the right order, on one small network.

Exercise L2.1

$R_{1} = 6 Ω$ is in series with the parallel combination of $R_{2} = 3 Ω$ and $R_{3} = 6 Ω$ . Find $R_{A B}$ .

Recall Solution L2.1

Step 1 — collapse the parallel pair (innermost group first). $R_{23} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2 Ω$ Why: $R_{2}, R_{3}$ share both nodes → same voltage. Step 2 — series add. $R_{A B} = R_{1} + R_{23} = 6 + 2 = 8 Ω$ Why: after collapsing, $R_{1}$ and the $2 Ω$ block share one lonely node → same current. ✓

Exercise L2.2

$R_{1} = 2 Ω$ in series with $R_{2} = 2 Ω$ ; that chain sits in parallel with $R_{3} = 2 Ω$ . Find $R_{A B}$ .

Recall Solution L2.2

Step 1 — series first (inner group). $R_{12} = 2 + 2 = 4 Ω$ Step 2 — parallel with $R_{3}$ . $R_{A B} = \frac{4 \cdot 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \approx 1.333 Ω$ Below the smallest branch ( $2 Ω$ )? Yes, $1.333 < 2$ . ✓

Exercise L2.3

Look at the figure. From A: $R_{1} = 4 Ω$ to node C. From C two branches go to B: $R_{2} = 6 Ω$ directly, and $R_{3} = 12 Ω$ directly. Find $R_{A B}$ .

Figure — Compute equivalent resistance in mixed networks

Recall Solution L2.2 → L2.3

Step 1 — the two branches C→B share nodes C and B → parallel. $R_{23} = \frac{6 \cdot 12}{6 + 12} = \frac{72}{18} = 4 Ω$ Step 2 — series with $R_{1}$ (node C now touches only $R_{1}$ and the merged block). $R_{A B} = 4 + 4 = 8 Ω$ ✓

Level 3 — Analysis

Goal: chain three or more collapses; forecast then verify.

Exercise L3.1 — the ladder

Terminals A–B. $R_{1} = 8 Ω$ from A to node C. From C: $R_{2} = 12 Ω$ straight to B, and $R_{3} = 4 Ω$ from C to D then $R_{4} = 8 Ω$ from D to B. Find $R_{A B}$ .

Recall Solution L3.1

Forecast: lower branch $R_{3} + R_{4} = 12 Ω$ in parallel with $R_{2} = 12 Ω$ → two equal paths → halves to $6 Ω$ ; then $+ 8$ . Guess ≈ $14 Ω$ . Step 1 — series in the lower branch (D touches only $R_{3}, R_{4}$ ): $R_{34} = 4 + 8 = 12 Ω$ Step 2 — parallel at C ( $R_{34}$ and $R_{2}$ share C and B): $R_{C} = \frac{12 \cdot 12}{12 + 12} = \frac{144}{24} = 6 Ω$ Step 3 — series with $R_{1}$ : $R_{A B} = 8 + 6 = 14 Ω$ Forecast (14) = verified (14). ✓

Exercise L3.2 — two parallel blocks in series

Between A and B: block 1 is ( $R_{1} = 10 Ω ∥ R_{2} = 10 Ω$ ), then in series with block 2 which is ( $R_{3} = 6 Ω ∥ R_{4} = 3 Ω$ ). Find $R_{A B}$ .

Recall Solution L3.2

Step 1 — first parallel block: $\frac{10 \cdot 10}{10 + 10} = 5 Ω$ . Step 2 — second parallel block: $\frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 Ω$ . Step 3 — the two blocks share one lonely node → series: $R_{A B} = 5 + 2 = 7 Ω$ ✓

Level 4 — Synthesis

Goal: strip away a disguise, spot short circuits and open branches.

Exercise L4.1 — the short-circuited resistor

Between A and B: $R_{1} = 5 Ω$ in series with $R_{2} = 10 Ω$ . A plain wire (0 Ω) is soldered directly across $R_{2}$ (from $R_{2}$ 's top node to $R_{2}$ 's bottom node). Find $R_{A B}$ .

Recall Solution L4.1

WHAT: the wire is a $0 Ω$ resistor in parallel with $R_{2}$ . $R_{2 ∥ 0} = \frac{R _{2} \cdot 0}{R _{2} + 0} = 0 Ω$ WHY it looks like this: every bit of current takes the free road (zero resistance), so no current flows through $R_{2}$ — it is shorted out and vanishes. Step — series left: $R_{A B} = R_{1} + 0 = 5 Ω$ Degenerate-case rule: a wire across a resistor deletes that resistor. ✓

Exercise L4.2 — the dangling (open) branch

Between A and B: $R_{1} = 7 Ω$ in series with $R_{2} = 3 Ω$ . A third resistor $R_{3} = 20 Ω$ is attached to the joint between $R_{1}$ and $R_{2}$ , but its far end connects to nothing (open circuit). Find $R_{A B}$ .

Recall Solution L4.2

WHAT: $R_{3}$ is a dead-end — a branch with an open end carries zero current (nowhere to return). WHY: by Kirchhoff's Current Law the current into a dangling node must be zero, so $R_{3}$ has no effect at all; it is as if it weren't there. Step: $R_{1}$ and $R_{2}$ are a clean series chain: $R_{A B} = 7 + 3 = 10 Ω$ Degenerate-case rule: an open (dangling) branch contributes nothing. ✓

Exercise L4.3 — balanced bridge (no Δ–Y needed!)

A Wheatstone bridge: from A, $R_{1} = 2 Ω$ to node C and $R_{2} = 4 Ω$ to node D. From C, $R_{3} = 4 Ω$ to B; from D, $R_{4} = 8 Ω$ to B. A bridging resistor $R_{5} = 100 Ω$ sits between C and D. Find $R_{A B}$ .

Recall Solution L4.3

Check the balance ratio first. A bridge is balanced when $\frac{R _{1}}{R _{3}} = \frac{R _{2}}{R _{4}}$ : $\frac{R _{1}}{R _{3}} = \frac{2}{4} = \frac{1}{2}, \frac{R _{2}}{R _{4}} = \frac{4}{8} = \frac{1}{2} .$ Equal → nodes C and D sit at the same voltage → no current crosses $R_{5}$ → the bridge resistor carries zero current and can be deleted (same logic as the open branch). We avoid a Delta-Wye Transformation entirely. Now it's plain series-parallel:

Top path A→C→B: $R_{1} + R_{3} = 2 + 4 = 6 Ω$ .
Bottom path A→D→B: $R_{2} + R_{4} = 4 + 8 = 12 Ω$ .
These two paths share A and B → parallel: $R_{A B} = \frac{6 \cdot 12}{6 + 12} = \frac{72}{18} = 4 Ω$ ✓

Level 5 — Mastery

Goal: a full multi-stage build, forecast, verify, and a downstream consequence.

Exercise L5.1 — the deep collapse

Terminals A–B. Trace the network:

$R_{1} = 4 Ω$ from A to node C.
From C, two branches to node E:
- Upper: $R_{2} = 6 Ω$ from C to D, then $R_{3} = 6 Ω$ from D to E.
- Lower: $R_{4} = 3 Ω$ directly from C to E.
$R_{5} = 5 Ω$ from E to B.

Find $R_{A B}$ , then use it to find the total current if a $9 V$ source drives A–B (a taste of Ohm's Law and the Current Divider).

Recall Solution L5.1

Forecast: upper branch $= 12 Ω$ in parallel with lower $3 Ω$ → below $3$ , roughly $2.4$ ; then $+ 4 + 5 =$ about $11 Ω$ . Step 1 — series in the upper branch (D touches only $R_{2}, R_{3}$ ): $R_{23} = 6 + 6 = 12 Ω$ Step 2 — parallel between C and E ( $R_{23}$ with $R_{4}$ ): $R_{C E} = \frac{12 \cdot 3}{12 + 3} = \frac{36}{15} = 2.4 Ω$ Step 3 — series chain A→C→E→B (all lonely joints now): $R_{A B} = R_{1} + R_{C E} + R_{5} = 4 + 2.4 + 5 = 11.4 Ω$ Forecast (≈11) ≈ verified (11.4). ✓

Downstream — total current from a $9 V$ source: $I = \frac{V}{R _{A B}} = \frac{9}{11.4} \approx 0.789 A$ Current split at node C (this whole current enters the C–E parallel pair). By the Current Divider, the fraction through a branch is proportional to the other branch's resistance: $I_{R_{4}} = I \cdot \frac{R _{23}}{R _{23} + R _{4}} = 0.789 \cdot \frac{12}{15} \approx 0.632 A$ $I_{upper} = I \cdot \frac{R _{4}}{R _{23} + R _{4}} = 0.789 \cdot \frac{3}{15} \approx 0.158 A$ Check: $0.632 + 0.158 = 0.789 A$ = the total. ✓ (Kirchhoff's Current Law holds.)

Active recall

Recall Test yourself before revealing

When you see a bridge, what do you check before doing any Δ–Y work? ::: The balance ratio $R_{1} / R_{3} = R_{2} / R_{4}$ ; if equal, the bridge resistor carries no current and can be deleted. A wire (0 Ω) across a resistor does what to that resistor? ::: Shorts it out — it carries no current and is deleted from the network. A resistor whose far end connects to nothing contributes what? ::: Nothing — an open branch carries zero current (KCL at a dead-end node). In a current divider, which branch gets more current? ::: The one with smaller resistance; each branch's share is weighted by the other branch's resistance. Why can't three parallel resistors use product-over-sum? ::: Product-over-sum is a two-resistor shortcut; for three, sum the conductances $1/ R$ and flip once.

Connections

Compute equivalent resistance in mixed networks — the parent note these exercises drill
Ohm's Law — turns $R_{A B}$ into a current in L5.1
Kirchhoff's Voltage Law — justifies every series add
Kirchhoff's Current Law — justifies every parallel add and the open-branch/short reasoning
Delta-Wye Transformation — needed only for an unbalanced bridge (L4.3 avoided it)
Current Divider — used in the L5.1 downstream split
Voltage Divider — the complementary rule (bigger R gets bigger voltage)
Conductance and Admittance — why parallel adds $1/ R$ , not $R$