Practice problems

Voltage divider practice problems

A voltage divider is two resistors in series, with the output taken from the point between them. It's how you scale a battery voltage down for a sensor input or set a reference. These ten problems start with the output voltage, then choose R2 for a target, and end with what happens when something is connected to the output.

The method

With R1 on top and R2 on the bottom, the output is the bottom resistor's share:

V_out = V_in × R2R1 + R2

Current through the divider: V_in over R1 + R2, with nothing connected to the output
Choosing R2: R2 = R1 × V_out over V_in − V_out
With a load: the divider acts like V_th (the unloaded output) behind R_th = R1 ∥ R2

The problems

Type your answer and tap Check. Every one has the working, step by step.

Problem 1 of 10
Two 10 kΩ resistors form a divider across 12 V. What's the output voltage?
Show the working
1. Equal resistors split the supply equally.
2. V_out = 12 V × 10 over 10 + 10 = 6 V
Answer: 6 V
Problem 2 of 10
A divider has R1 = 2.2 kΩ on top and R2 = 1 kΩ on the bottom, across 12 V. What's the output voltage?
Show the working
1. Use V_out = V_in × R2 over R1 + R2, with the bottom resistor on top of the fraction.
2. V_out = 12 V × 1 over 2.2 + 1 = 12 V × 1 over 3.2 = 3.75 V
Answer: 3.75 V
Problem 3 of 10
A divider has R1 = 10 kΩ on top and R2 = 4.7 kΩ on the bottom, across 5 V. What's the output voltage?
Show the working
1. V_out = 5 V × 4.7 over 10 + 4.7 = 5 V × 4.7 over 14.7
2. V_out ≈ 1.60 V. The smaller resistor is on the bottom, so the output is less than half.
Answer: 1.6 V
Problem 4 of 10
A 4.7 kΩ and a 3.3 kΩ resistor in series make a divider across 12 V. How much current flows through the divider, in milliamps?
Show the working
1. Nothing is connected to the output, so the divider is just two resistors in series: 4.7 kΩ + 3.3 kΩ = 8 kΩ.
2. I = 12 V over 8 kΩ = 1.5 mA
Answer: 1.5 mA
Problem 5 of 10
A divider has R1 = 4.7 kΩ on top and R2 = 10 kΩ on the bottom, across 6 V. What voltage does the top resistor drop?
Show the working
1. The top resistor's share is R1 over R1 + R2 of the supply: 6 V × 4.7 over 14.7 ≈ 1.92 V.
2. Check with KVL: the output is 6 V × 10 over 14.7 ≈ 4.08 V, and 6 − 4.08 = 1.92 V.
Answer: 1.92 V
Problem 6 of 10
You want 3 V from a 9 V supply. R1 on top is 10 kΩ. What value does R2 need to be?
Show the working
1. 3 V out of 9 V leaves 6 V across R1. In series, the voltages split in the same ratio as the resistances.
2. R2 = R1 × V_out over V_in − V_out = 10 kΩ × 3 over 6 = 5 kΩ. The nearest standard values are 4.7 kΩ and 5.6 kΩ.
Answer: 5 kΩ
Problem 7 of 10
You want 5 V from a 12 V supply, with R1 = 22 kΩ on top. What value of R2 gives exactly 5 V?
Show the working
1. R1 takes 12 − 5 = 7 V, so R2 = 22 kΩ × 5 over 7.
2. R2 ≈ 15.7 kΩ. The nearest E12 value, 15 kΩ, gives 12 V × 15 over 37 ≈ 4.86 V.
Answer: 15.7 kΩ
Problem 8 of 10
A divider has R1 = 10 kΩ on top and R2 = 15 kΩ on the bottom, across 5 V. A 30 kΩ load is connected from the output to ground. What's the output voltage now?
Show the working
1. Without the load, the output would be 5 V × 15 over 25 = 3 V. The load sits in parallel with R2: 15 × 30 over 15 + 30 = 10 kΩ.
2. Now the divider is 10 kΩ over 10 kΩ: V_out = 5 V × 10 over 10 + 10 = 2.5 V. The load pulled it down by 0.5 V.
Answer: 2.5 V
Problem 9 of 10
A divider has R1 = 2.2 kΩ on top and R2 = 3.3 kΩ on the bottom, across 5 V. What's the output with a 22 kΩ load connected?
Show the working
1. With nothing connected: V_th = 5 V × 3.3 over 5.5 = 3 V.
2. Seen from the output, the divider is 3 V behind R_th = 2.2 kΩ ∥ 3.3 kΩ = 2.2 × 3.3 over 5.5 = 1.32 kΩ.
3. The load makes a second divider: V = 3 V × 22 over 1.32 + 22 ≈ 2.83 V
Answer: 2.83 V
Problem 10 of 10
Two 10 kΩ resistors divide 5 V and feed a sensor with a 100 kΩ input. By what percentage does the sensor's voltage sag below the unloaded 2.5 V?
Show the working
1. R_th = 10 kΩ ∥ 10 kΩ = 5 kΩ, so the sensor gets 2.5 V × 100 over 5 + 100 ≈ 2.381 V.
2. The sag is 2.5 − 2.381 over 2.5, which is the same as R_th over R_th + R_L = 5 over 105.
3. That's 4.76%, close to the quick estimate R_th over R_L = 5%.
Answer: 4.76%

Build it in a free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

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Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Voltage dividers” is lesson 7 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

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The ideas behind it

Questions people ask

What is the voltage divider formula?

V_out = V_in × R2 ÷ (R1 + R2), where R2 is the bottom resistor, between the output and ground. It's the bottom resistor's share of the total. With equal resistors, the output is exactly half the supply.

How do I choose resistors for a voltage divider?

Pick R1, then R2 = R1 × V_out ÷ (V_in − V_out). Only the ratio sets the output, so choose the size by the current you can spare and the load: R1 and R2 in parallel should be about ten times smaller than the load's resistance.

Why does my divider's output drop when I connect something?

The load sits in parallel with R2, which lowers the bottom resistance and so the output. This is called loading. Seen from its output, the divider acts like V_th behind R1 ∥ R2, and the output sags by about R_th ÷ R_L.