Build and analyze a voltage divider: Exercises — Hardware deep dive

Reminders you will reuse (all from the parent):

Same current through both: $I = \frac{V _{in}}{R _{1} + R _{2}}$ (this is Ohm's Law on the whole loop).
The two drops add back to the source: $V_{R_{1}} + V_{R_{2}} = V_{in}$ (this is Kirchhoff's Voltage Law).
A load $R_{L}$ hung across $R_{2}$ replaces $R_{2}$ with the parallel combination $R_{2} ∥ R_{L} = \frac{R _{2} R _{L}}{R _{2} + R _{L}}$ (see Series and Parallel Resistors).
The divider's Thevenin output resistance is $R_{t h} = R_{1} ∥ R_{2}$ (see Thevenin Equivalent).

Figure — Build and analyze a voltage divider

The picture above is the reference circuit for every problem: input at top, $R_{1}$ then $R_{2}$ going down to ground, and the output tapped at the joint between them.

Level 1 — Recognition

Can you read the circuit and plug in?

Exercise 1.1

$V_{in} = 10 V$ , $R_{1} = R_{2} = 4.7 k Ω$ . Find $V_{o u t}$ .

Recall Solution 1.1

WHAT: equal resistors. WHY the shortcut: if $R_{1} = R_{2}$ , the fraction is $\frac{R _{2}}{R _{1} + R _{2}} = \frac{1}{2}$ , so the input splits exactly in half — no arithmetic on the $4.7$ needed. $V_{o u t} = 10 \cdot \frac{4700}{4700 + 4700} = 10 \cdot \frac{1}{2} = 5 V$ Answer: $V_{o u t} = 5$ V.

Exercise 1.2

$V_{in} = 12 V$ , $R_{1} = 1 k Ω$ , $R_{2} = 3 k Ω$ . Find $V_{o u t}$ .

Recall Solution 1.2

WHAT: plug straight in. WHY: $R_{2}$ owns $3$ of the $4$ total kilo-ohms, so it should keep three-quarters of the voltage. $V_{o u t} = 12 \cdot \frac{3000}{1000 + 3000} = 12 \cdot \frac{3}{4} = 9 V$ Answer: $V_{o u t} = 9$ V. Sanity: $R_{2}$ is the bigger resistor, so it drops the bigger voltage — matches the "bigger resistor = bigger drop" rule.

Exercise 1.3

Same circuit as 1.2. Without a new formula, find $V_{R_{1}}$ (the voltage across $R_{1}$ ).

Recall Solution 1.3

WHAT: use KVL instead of a second formula. WHY: the two drops must add to $V_{in}$ , so once we know $V_{o u t}$ we get the other drop by subtraction. $V_{R_{1}} = V_{in} - V_{o u t} = 12 - 9 = 3 V$ Check with the twin formula $V_{R_{1}} = V_{in} \frac{R _{1}}{R _{1} + R _{2}} = 12 \cdot \frac{1000}{4000} = 3$ V. ✓ Answer: $V_{R_{1}} = 3$ V.

Level 2 — Application

Now run the current and drops all the way through.

Exercise 2.1

$V_{in} = 9 V$ , $R_{1} = 2.2 k Ω$ , $R_{2} = 3.3 k Ω$ . Find (a) the loop current $I$ , (b) $V_{o u t}$ , (c) confirm KVL.

Recall Solution 2.1

(a) Current. WHY first: current is shared by both resistors, so it's the natural anchor. Using Ohm's Law on the whole loop: $I = \frac{V _{in}}{R _{1} + R _{2}} = \frac{9}{2200 + 3300} = \frac{9}{5500} \approx 1.636 mA$ (b) Output. Ohm's law on $R_{2}$ alone: $V_{o u t} = I \cdot R_{2} = 1.636 mA \times 3300 Ω \approx 5.4 V$ Cross-check with the divider formula: $9 \cdot \frac{3300}{5500} = 5.4$ V. ✓ (c) KVL. $V_{R_{1}} = I \cdot R_{1} = 1.636 mA \times 2200 = 3.6$ V, and $3.6 + 5.4 = 9 = V_{in}$ . ✓ Answers: $I \approx 1.636$ mA, $V_{o u t} = 5.4$ V, $V_{R_{1}} = 3.6$ V.

Exercise 2.2 (design a ratio)

You want $V_{o u t} = 1.8 V$ from $V_{in} = 5 V$ . If you fix $R_{2} = 10 k Ω$ , what $R_{1}$ do you need?

Recall Solution 2.2

WHAT: solve the divider equation for $R_{1}$ . WHY $R_{2}$ can be fixed freely: only the ratio sets $V_{o u t}$ , so we're free to pin one resistor and solve for the other. $\frac{V _{o u t}}{V _{in}} = \frac{R _{2}}{R _{1} + R _{2}} \Rightarrow \frac{1.8}{5} = \frac{10000}{R _{1} + 10000}$ $R_{1} + 10000 = \frac{10000}{0.36} = 27777.8 \Rightarrow R_{1} = 17777.8 Ω \approx 17.8 k Ω$ Answer: $R_{1} \approx 17.8 k Ω$ . Check: $5 \cdot \frac{10000}{27777.8} = 1.8$ V. ✓

Exercise 2.3 (power)

For the circuit in 2.1, how much power does $R_{1}$ dissipate?

Recall Solution 2.3

WHAT: power in a resistor is $P = I^{2} R$ (or $V^{2} / R$ ). WHY $I^{2} R$ here: we already know the exact current through $R_{1}$ , so squaring it is one clean step. $P_{R_{1}} = I^{2} R_{1} = (1.636 \times 1 0^{- 3})^{2} \times 2200 \approx 5.89 \times 1 0^{- 3} W \approx 5.89 mW$ Cross-check with $V_{R_{1}}^{2} / R_{1} = 3. 6^{2} /2200 \approx 5.89$ mW. ✓ Answer: $\approx 5.89$ mW — tiny, so a standard $\frac{1}{4}$ -W resistor is fine.

Level 3 — Analysis

Explain the "why," handle loading, edge cases.

Exercise 3.1 (loading effect)

$V_{in} = 12 V$ , $R_{1} = R_{2} = 4.7 k Ω$ (unloaded $V_{o u t} = 6$ V). You connect a load $R_{L} = 4.7 k Ω$ across the output. Find the new $V_{o u t}$ and explain the drop.

Recall Solution 3.1

WHAT: the load sits in parallel with $R_{2}$ because both span the same two nodes (output and ground). WHY parallel: two elements sharing both endpoints always combine as a parallel pair (see Series and Parallel Resistors). $R_{2} ∥ R_{L} = \frac{R _{2} R _{L}}{R _{2} + R _{L}} = \frac{4700 \cdot 4700}{4700 + 4700} = 2350 Ω$ Now re-run the divider with this smaller bottom resistance: $V_{o u t} = 12 \cdot \frac{2350}{4700 + 2350} = 12 \cdot \frac{2350}{7050} = 4 V$ Answer: $V_{o u t} = 4$ V (down from 6 V). WHY it dropped: the load "steals" current away, so the effective bottom resistor shrank from $4.7$ k to $2.35$ k, and a smaller bottom share means a smaller voltage share.

Exercise 3.2 (Thevenin view)

For the unloaded divider in 3.1, find the Thevenin equivalent seen by any load: the open-circuit voltage $V_{t h}$ and the output resistance $R_{t h}$ .

Recall Solution 3.2

WHAT: replace the whole divider by one source $V_{t h}$ behind one resistor $R_{t h}$ (this is the Thevenin Equivalent).

$V_{t h} =$ the unloaded output $= 6$ V (open circuit means no load current, so nothing droops).
$R_{t h} = R_{1} ∥ R_{2}$ : WHY — to find the resistance looking back, kill the source (short $V_{in}$ ), and then $R_{1}$ and $R_{2}$ both dangle from the output node to ground in parallel. $R_{t h} = \frac{4700 \cdot 4700}{9400} = 2350 Ω$ Answer: $V_{t h} = 6$ V, $R_{t h} = 2350 Ω = 2.35 k Ω$ . Bonus consistency check: load with $R_{L} = 4.7$ k and this Thevenin model predicts $6 \cdot \frac{4700}{2350 + 4700} = 6 \cdot \frac{4700}{7050} = 4$ V — exactly Exercise 3.1. Two methods, same answer. ✓

Exercise 3.3 (degenerate cases)

Reason out $V_{o u t}$ (no calculator) for each edge case of $R_{1} = R_{2}$ -style circuits: (a) $R_{2} = 0$ ; (b) $R_{1} = 0$ ; (c) $R_{2} \to \infty$ (open circuit) with $V_{in} = 8$ V throughout.

Recall Solution 3.3

Work each from the fraction $\frac{R _{2}}{R _{1} + R _{2}}$ and picture the circuit, don't just plug in.

(a) $R_{2} = 0$ (bottom is a plain wire): $\frac{0}{R _{1} + 0} = 0$ , so $V_{o u t} = 0$ V. WHY: the output node is wired straight to ground — a wire cannot hold any voltage across itself.
(b) $R_{1} = 0$ (top is a plain wire): $\frac{R _{2}}{0 + R _{2}} = 1$ , so $V_{o u t} = V_{in} = 8$ V. WHY: nothing drops the voltage before the tap, so the full input reaches the output.
(c) $R_{2} \to \infty$ (output left open / broken): $\frac{R _{2}}{R _{1} + R _{2}} \to 1$ , so $V_{o u t} \to V_{in} = 8$ V. WHY: with no path to ground through $R_{2}$ , no current flows, so there's zero drop across $R_{1}$ (Ohm: $V = I R = 0 \cdot R_{1}$ ), and the output floats up to the full input. Answers: (a) 0 V, (b) 8 V, (c) 8 V. Notice (b) and (c) give the same number for opposite reasons — one drops nothing on top, the other draws no current at all.

Level 4 — Synthesis

Combine constraints: voltage AND current AND power.

Exercise 4.1 (design under a current budget)

Make $V_{o u t} = 3.3 V$ from $V_{in} = 5 V$ , and the standing current through the divider must be at least $0.5 mA$ (so a light $10 k Ω$ sensor load barely disturbs it). Choose $R_{1}$ and $R_{2}$ .

Recall Solution 4.1

Step 1 — the ratio (voltage constraint). $\frac{R _{2}}{R _{1} + R _{2}} = \frac{3.3}{5} = 0.66$ . Step 2 — the total resistance (current constraint). Current is set by the total: $I = \frac{V _{in}}{R _{1} + R _{2}} \geq 0.5 mA \Rightarrow R_{1} + R_{2} \leq \frac{5}{0.5 \times 1 0 ^{- 3}} = 10000 Ω$ WHY two separate constraints: the ratio fixes the split (voltage); the total fixes the current. They're independent knobs. Step 3 — pick the total at the limit $R_{1} + R_{2} = 10 k Ω$ to keep current at exactly $0.5$ mA (minimizes power while meeting the budget): $R_{2} = 0.66 \times 10000 = 6.6 k Ω, R_{1} = 10000 - 6600 = 3.4 k Ω$ Answer: $R_{1} = 3.4 k Ω, R_{2} = 6.6 k Ω$ , giving $I = 0.5$ mA and $V_{o u t} = 5 \cdot \frac{6600}{10000} = 3.3$ V. ✓

Exercise 4.2 (stiffness check on your own design)

Take the divider from 4.1 and attach the $R_{L} = 10 k Ω$ sensor load. How far does $V_{o u t}$ actually move? Is the design "stiff enough" (error under 10%)?

Recall Solution 4.2

Step 1 — parallel the load onto $R_{2}$ : $R_{2} ∥ R_{L} = \frac{6600 \cdot 10000}{6600 + 10000} = \frac{66 000 000}{16600} \approx 3975.9 Ω$ Step 2 — re-run the divider: $V_{o u t} = 5 \cdot \frac{3975.9}{3400 + 3975.9} = 5 \cdot \frac{3975.9}{7375.9} \approx 2.695 V$ Step 3 — error: $\frac{3.3 - 2.695}{3.3} \approx 0.183 = 18.3%$ . Answer: $V_{o u t} \approx 2.70$ V, an 18% sag — NOT stiff enough. WHY it failed the budget promise: "0.5 mA standing current" isn't automatically $≫$ the load current; here the load is only $10$ k against an output impedance $R_{t h} = R_{1} ∥ R_{2} = 3.4 ∥ 6.6 \approx 2.24$ k, and $R_{L}$ is not $≫ R_{t h}$ . Fix: shrink both resistors $\sim 10 \times$ (keep the ratio) or add an Op-amp Buffer.

Level 5 — Mastery

Two dividers, a bridge, and a general proof.

Exercise 5.1 (Wheatstone-style comparison)

Two dividers share the same $V_{in} = 6 V$ . Left leg: $R_{1} = 1 k Ω$ (top), $R_{2} = 1 k Ω$ (bottom), output node $A$ . Right leg: $R_{3} = 2 k Ω$ (top), $R_{4} = 3 k Ω$ (bottom), output node $B$ . Find $V_{A}$ , $V_{B}$ , and the bridge voltage $V_{A} - V_{B}$ (this is a Wheatstone Bridge).

Recall Solution 5.1

WHAT: each leg is an independent voltage divider; the bridge voltage is the difference of their taps. Left tap: $V_{A} = 6 \cdot \frac{R _{2}}{R _{1} + R _{2}} = 6 \cdot \frac{1000}{2000} = 3 V$ . Right tap: $V_{B} = 6 \cdot \frac{R _{4}}{R _{3} + R _{4}} = 6 \cdot \frac{3000}{2000 + 3000} = 6 \cdot \frac{3}{5} = 3.6 V$ . Bridge: $V_{A} - V_{B} = 3 - 3.6 = - 0.6 V$ . Answer: $V_{A} = 3$ V, $V_{B} = 3.6$ V, $V_{A} - V_{B} = - 0.6$ V. WHY the sign matters: $B$ sits higher than $A$ , so a meter reading $V_{A} - V_{B}$ shows negative — the bridge is unbalanced, and the sign tells you which leg has the larger bottom share. Balance ( $V_{A} - V_{B} = 0$ ) needs $\frac{R _{2}}{R _{1} + R _{2}} = \frac{R _{4}}{R _{3} + R _{4}}$ , i.e. equal ratios.

Exercise 5.2 (balance condition — cross-multiply)

For the bridge above, keep $R_{1} = R_{2} = 1 k Ω$ and $R_{3} = 2 k Ω$ . What value of $R_{4}$ balances the bridge ( $V_{A} = V_{B}$ )?

Recall Solution 5.2

WHAT: set the two tap ratios equal. WHY: balance means both taps hit the same voltage, and each tap is purely its own ratio. $\frac{R _{2}}{R _{1} + R _{2}} = \frac{R _{4}}{R _{3} + R _{4}} \Rightarrow \frac{1000}{2000} = \frac{R _{4}}{2000 + R _{4}}$ $\frac{1}{2} (2000 + R_{4}) = R_{4} \Rightarrow 1000 + \frac{1}{2} R_{4} = R_{4} \Rightarrow 1000 = \frac{1}{2} R_{4} \Rightarrow R_{4} = 2000 Ω$ Answer: $R_{4} = 2 k Ω$ . Check: right tap $= 6 \cdot \frac{2000}{4000} = 3$ V $= V_{A}$ . ✓ Reading it: with $R_{1} = R_{2}$ the left ratio is $\frac{1}{2}$ ; the right leg balances only when its ratio is also $\frac{1}{2}$ , i.e. $R_{4} = R_{3}$ .

Exercise 5.3 (general proof — why only the ratio matters)

Prove that scaling both resistors by the same factor $k > 0$ leaves $V_{o u t}$ unchanged. Then state the one thing that does change.

Recall Solution 5.3

Claim: replacing $R_{1} \to k R_{1}$ and $R_{2} \to k R_{2}$ gives the same $V_{o u t}$ . $V_{o u t}^{'} = V_{in} \cdot \frac{k R _{2}}{k R _{1} + k R _{2}} = V_{in} \cdot \frac{k R _{2}}{k ( R _{1} + R _{2} )} = V_{in} \cdot \frac{R _{2}}{R _{1} + R _{2}} = V_{o u t}$ WHY: the common factor $k$ cancels top and bottom — the ratio is scale-invariant, which is exactly why "only the ratio matters" for the voltage. $■$ What DOES change — the current (and power): $I^{'} = \frac{V _{in}}{k R _{1} + k R _{2}} = \frac{1}{k} \cdot \frac{V _{in}}{R _{1} + R _{2}} = \frac{I}{k}$ Scaling up by $k$ divides the current by $k$ and cuts total power by $k$ (lower waste), but raises output impedance $R_{t h} = k (R_{1} ∥ R_{2})$ by $k$ — worse loading. That's the exact trade-off behind "bigger resistors aren't a free win."

Wrap-up recall

Voltage divider ratio for equal resistors

exactly

\frac{1}{2}

, so output is half the input.

To find

V_{R_{1}}

quickly once you know

V_{o u t}

use KVL:

V_{R_{1}} = V_{in} - V_{o u t}

Loading rule for a stiff output

need

R_{L} ≫ R_{t h} = R_{1} ∥ R_{2}

(about

10 \times

Divider output impedance

R_{t h} = R_{1} ∥ R_{2}

What scaling both resistors by

k

changes

not

V_{o u t}

; it divides current by

k

and multiplies

R_{t h}

k

Bridge balance condition

equal tap ratios

\frac{R _{2}}{R _{1} + R _{2}} = \frac{R _{4}}{R _{3} + R _{4}}

Connections

Parent: 1.2.5 Voltage divider — the formula and derivation these exercises drill.
Ohm's Law — every current/drop step is Ohm's law.
Series and Parallel Resistors — series adds; loading brings in parallel.
Kirchhoff's Voltage Law — drops sum to source (used in L1/L2 checks).
Thevenin Equivalent — $V_{t h}$ , $R_{t h}$ in L3/L4.
Wheatstone Bridge — two dividers compared in L5.
Op-amp Buffer — the fix when loading is too severe.