Graded voltage divider and Wheatstone bridge labs: design a divider, see loading, make it stiff, balance a bridge and read an unbalanced one.
On the syllabus
The voltage divider rule
Loading and output resistance (Thevenin view of a divider)
The Wheatstone bridge, balanced and unbalanced
Measuring an unknown resistance
You will be able to
Design a divider for a given output voltage.
Predict how a load pulls a divider down, and design one that holds up under load.
Balance a Wheatstone bridge and find an unknown from the balance condition.
Read the output of an unbalanced bridge as a difference of two dividers.
The first challenge is free. The rest, the viva and the certificate are part of Pro, and everything is here to read.
Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Design a divider for 3.3 V
Two resistors in series split a voltage. The output is a fixed fraction of the input, set by the ratio of the resistors, not their size.
Worked example9 V down to 3 V with R1 = 12 kΩ: R2 = 12 kΩ × 3 / 6 = 6 kΩ. The nearest E24 value, 6.2 kΩ, gives 9 × 6.2 / 18.2 = 3.07 V.
On the scope or meterA voltmeter on the output, black on ground. The reading changes when you change the ratio of R1 and R2.
Your task
R1 is 10 kΩ from 5 V to the output. Choose R2, from the output to ground, so the output is 3.3 V within 2%.
Use Vout = Vin × R2 / (R1 + R2), then pick a standard value.
2. See a load pull a divider down
A load on a divider sits in parallel with the bottom resistor, so the output drops. Another view: a divider is a Thevenin source, a voltage Vth behind a resistance Rth = R1 ∥ R2.
FormulaVout = Vth × RL / (Rth + RL)
Worked example12 V across two 20 kΩ resistors: Vth = 6 V and Rth = 10 kΩ. A 10 kΩ load gives 6 × 10 / 20 = 3 V, half of what it read unloaded.
On the scope or meterThe output falls the moment the load connects. The load current is the output voltage divided by RL.
Your task
Two 10 kΩ resistors divide 10 V. With nothing on the output it would read 5 V.
A 10 kΩ load, RL, hangs on the output. Measure the output voltage and the current in the load.
3. Make a divider stiff enough for its load
A divider holds up under load only if its output resistance is small next to the load. Smaller resistors make it stiffer, but they draw more current.
FormulaVout = Vin × (R2 ∥ RL) / (R1 + R2 ∥ RL)
Worked example10 V into 5 V for a 22 kΩ load, with R1 = 2.2 kΩ: R2 ∥ RL must be 2.2 kΩ, so R2 = 2.2 × 22 / (22 − 2.2) = 2.44 kΩ. 2.4 kΩ gives 4.96 V, and R1 carries 2.3 mA.
On the scope or meterThe output with the load connected, and the current in R1. Both resistors change the output; only R1 sets the current limit.
Your task
Divide 12 V for a 10 kΩ load, RL. With the load on, the output must be 6 V within 3%, and R1 must carry no more than 10 mA.
Choose R1 and R2. Hint: the load is in parallel with R2, so it is R2 in parallel with RL that has to match R1.
4. Balance a Wheatstone bridge
A Wheatstone bridge is two dividers side by side. It balances when both midpoints sit at the same voltage, and that happens when the two ratios match.
FormulaR1 / R2 = R3 / Rx, so Rx = R2 × R3 / R1
Worked exampleR1 = 2 kΩ, R2 = 5 kΩ and R3 = 4 kΩ: Rx = 5 × 4 / 2 = 10 kΩ. Any supply voltage gives zero across the middle.
On the scope or meterA voltmeter from one midpoint to the other. It reads zero at balance, and its sign flips as Rx passes the balance value.
Your task
The left arm is R1 (1 kΩ) over R2 (2.2 kΩ); the right arm is R3 (1 kΩ) over Rx. Both arms run from 10 V to ground.
Set Rx so the bridge balances: the voltage between the two midpoints within 20 mV of zero.
5. Read an unbalanced bridge
An unbalanced bridge gives a small voltage: the difference between two divider outputs. A small change in one arm shows up clearly because the large equal parts cancel.
On the scope or meterRed lead on the right midpoint, black on the left one, in millivolts.
Your task
Same bridge, all arms 1 kΩ except Rx, which is now 1.2 kΩ: a strain gauge that has stretched, say.
Measure the bridge output: the right midpoint minus the left one, in millivolts. Predict it first as the difference of two dividers.
6. Maximum power transfer
Thevenin’s theorem: any network of sources and resistors, seen from two terminals, acts like one voltage Vth behind one resistance Rth. Vth is the open-circuit voltage. Rth is what you see looking back with the sources shorted. A load gets the most power when it equals Rth.
FormulaPmax = Vth² / (4 Rth), at RL = Rth
Worked exampleVth = 6 V and Rth = 2 kΩ: at RL = 2 kΩ the load gets 36 / 8000 = 4.5 mW. At RL = 4 kΩ, I = 6 / 6000 = 1 mA and P = 4 mW: less.
On the scope or meterThe power in RL rises, peaks and falls as RL grows. At the peak, RL has exactly half of Vth across it.
Your task
A 12 V supply feeds a 6 kΩ / 3 kΩ divider, and a 1 kΩ resistor, R3, runs from the divider’s middle to the load RL.
Choose RL so it takes as much power as this network can give: at least 1.31 mW. Find the Thevenin equivalent the load sees first.
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.