Voltage Divider Practice Problems (Loading, ADC Input) with Solutions

A divider sets a voltage by a ratio. The key point is that a load changes that ratio.

Key formulas

V_out = V_in × (R₂)/(R₁ + R₂)

R₂ = R₁ × (V_out)/(V_in − V_out)

R_TH = R₁ ∥ R₂

Problems 9

3 multiple-choice and 6 numeric problems. Type numbers in the unit shown next to the box (for example 2.55); SI prefixes such as 4.7k also work. Small differences from rounding along the way count as correct (±2%). When you are done, press Submit and grade at the bottom to see the answers and step-by-step solutions.

A divider has 12 V in, R1 = 10 kΩ (top) and R2 = 4.7 kΩ (bottom). What is the unloaded output voltage, in volts?
V
Answer: 3.837 V
Solution
1. The output is across the bottom resistor R2: V_out = V_in × (R₂)/(R₁ + R₂).
2. V_out = 12 × (4.7)/(14.7) = 12 × 0.3197 = 3.837 V
3. Putting R1 on top of the fraction is a common slip: the resistor the output is taken across goes on top.
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A divider drops 5 V to 3.3 V and R1 = 10 kΩ. What should R2 be, in kΩ?
kΩ
Answer: 19.41 kΩ
Solution
1. Solving the divider formula for R2 gives R₂ = R₁ × (V_out)/(V_in − V_out).
2. R₂ = 10 kΩ × (3.3)/(1.7) = 19.41 kΩ
3. The nearest standard value is 20 kΩ, which gives 3.33 V.
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From a 5 V input, which (R1, R2) pair gives the output closest to 3.3 V?

AR1 = 10 kΩ, R2 = 10 kΩBR1 = 10 kΩ, R2 = 20 kΩCR1 = 4.7 kΩ, R2 = 10 kΩDR1 = 1 kΩ, R2 = 2.2 kΩ
Answer: B. R1 = 10 kΩ, R2 = 20 kΩ
Solution
1. Work out each pair with V_out = 5 × (R₂)/(R₁ + R₂).
2. (10, 10) → 2.5 V, (10, 20) → 3.33 V, (4.7, 10) → 3.4 V, (1, 2.2) → 3.44 V
3. The smallest gap to 3.3 V is R1 = 10 kΩ, R2 = 20 kΩ.
Check it with the Voltage divider calculator
A divider with 10 V in and R1 = R2 = 10 kΩ drives a 10 kΩ load. What is the output voltage, in volts?
V
Answer: 3.333 V
Solution
1. The load is in parallel with R2: 10 kΩ ∥ 10 kΩ = 5 kΩ.
2. V_out = 10 × (5)/(10 + 5) = 3.333 V
3. That is 1.67 V below the unloaded 5 V. Keep the load at least ten times the divider resistors for a small error.
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What is the main reason not to power a small motor from a voltage divider?

AResistors are too expensiveBThe output sags and varies once load current flowsCDividers only work with ACDThe output rises above the input
Answer: B. The output sags and varies once load current flows
Solution
1. The output changes as soon as a load sits in parallel with R2. A motor draws a large, changing current, so the voltage swings widely.
2. Use a regulator for power, and keep dividers for inputs that draw almost no current, such as an ADC pin.
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A divider has 24 V in, R1 = 47 kΩ and R2 = 10 kΩ, with no load. How much current does it draw from the supply, in mA?
mA
Answer: 0.4211 mA
Solution
1. With no load the two resistors are simply in series: 47 + 10 = 57 kΩ.
2. I = (24 V)/(57 kΩ) = 0.4211 mA
3. In battery equipment this current flows all the time, so divider resistors are often made large to cut it.
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You want to measure up to 15 V with a 3.3 V-reference ADC. With R2 = 10 kΩ, what R1 makes the output exactly 3.3 V at 15 V, in kΩ?
kΩ
Answer: 35.45 kΩ
Solution
1. Use R₁ = R₂ × ((V_in)/(V_out) − 1).
2. R₁ = 10 kΩ × ((15)/(3.3) − 1) = 10 × 3.545 = 35.45 kΩ
3. In practice pick a slightly larger standard value (such as 39 kΩ) so 15 V stays safely below 3.3 V.
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What is the Thévenin (output) resistance of a divider with R1 = 10 kΩ and R2 = 4.7 kΩ, in kΩ?
kΩ
Answer: 3.197 kΩ
Solution
1. With the source shorted, the output sees R1 and R2 in parallel.
2. R_TH = (10 × 4.7)/(10 + 4.7) = (47)/(14.7) = 3.197 kΩ
3. A load forms another divider with this resistance, so the load must be much larger than it.
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Both R1 and R2 of an unloaded divider are made ten times larger. What changes?

AThe output voltage rises tenfoldBThe output stays the same and the current drops to a tenthCThe output drops to a tenthDNeither the output nor the current changes
Answer: B. The output stays the same and the current drops to a tenth
Solution
1. The output depends only on the ratio (R₂)/(R₁ + R₂), so scaling both by the same factor leaves it unchanged.
2. The total resistance grows tenfold, so the current falls to a tenth. The Thévenin resistance also grows tenfold, so loads pull the output down more.
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