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CR · Voltage division

The Voltage Divider

How do two resistors in series share the supply voltage? One idea is enough: the same current flows through both, so the voltage divides in direct proportion to resistance. You learn why the output takes R2's share of the supply voltage from a picture in which the resistance bar and the voltage bar move by exactly the same length.

Same current, voltage splits in the resistance ratio

V_in = 12V feeds R1 = 4kΩ and a second resistor R2 in series, and the output is taken across R2. Moving R2 changes the shares of resistance split between R1 and R2 in the top bar, and the bottom bar is the output voltage, the part of the supply that matches that share. The length of the right-hand R2 segment in the top bar always equals the length of the output bar below. Set R2 so the output reaches the target of 8V on the orange dashed line.

Second resistance R2R2 = 2.0 kΩ
Output voltage and divider ratio
V_out = 4.0 V · R2/(R1+R2) = 0.33
Far from 8V

The same current passes through both resistors

A series connection is one path with no branches, so the same current I = V_in/(R1+R2) passes through the two resistors in turn. By Ohm's law the voltages across them are V1 = I·R1 and V2 = I·R2. Since the current is identical in both, the voltage ratio V2/V1 equals the resistance ratio R2/R1, and the larger resistor takes the larger voltage. That single line, voltage proportional to resistance, is the whole of voltage division.

The output is R2’s share

Take the output across R2 and V_out = V2 = I·R2; substitute I = V_in/(R1+R2) and V_out = V_in·R2/(R1+R2). The division ratio R2/(R1+R2) is simply the fraction of the total resistance that R2 makes up, so when R2 is much larger than R1 the ratio nears 1 and the output approaches the supply voltage, and when R2 is small the ratio nears 0 and the output almost vanishes. Raising R2, then, means raising R2's share of the voltage.

A tool to read, and the mirror of current division

Voltage division is the basic tool for making a reference voltage or for scaling a large voltage down to a size a meter can read. But the formula holds as written only while the load current drawn from the output is small. Attach a load to the output and it sits in parallel with the output-side resistor, effectively shrinking the lower resistance, so the output voltage falls below the formula's value; that is why exam problems treat the unloaded divider as the default. Within the same series-parallel analysis, current division, where parallel branches share current in the ratio of their conductances, is the mirror of the voltage divider. Series splits voltage by resistance; parallel splits current by conductance.

ObserveI = Vin(R1+R2)
In series, the current is the same.
ChooseVout = ?
The output is the voltage across R2.
Fill inVout = ?
The ratio is the output-side resistance over the total.
On your ownVoutVin = ?
Voltage splits in the resistance ratio.

Back to the first screen

As you raised R2 the output bar below grew, and it reached the target of 8V when R2 was 8kΩ (R1 = 4, V_in = 12). The segment R2 occupied in the top resistance bar and the length of the output bar below matched from start to finish, because the same current ran through both resistors and the voltage split in direct proportion to resistance. The division ratio R2/(R1+R2), the share of the total resistance that R2 takes, sets the output voltage.

Two resistors in series carry the same current, so they split the supply voltage in the resistance ratio. The output across R2 is Vout = Vin·R2/(R1+R2), and the division ratio R2/(R1+R2) is R2's share of the total resistance. This holds while the load current at the output is small, and current division, which splits current in parallel by the conductance ratio, is its mirror.