Voltage Divider

Two resistors in series can produce a fraction of the input voltage. R1 is the upper resistor, and R2 is the lower resistor. The output is the midpoint voltage measured relative to ground, so it equals the voltage across R2.

Here a $6\,\mathrm{V}$ source drives $R_1=1\,\mathrm{k}\Omega$ and $R_2=2\,\mathrm{k}\Omega$. With no load connected to the midpoint:

$$\begin{aligned} V_{\mathrm{out}} &= V_{\mathrm{in}}\frac{R_2}{R_1+R_2}\\ &= 6\frac{2}{1+2}=4\,\mathrm{V} \end{aligned}$$

Use the lower resistor in the numerator when measuring the midpoint relative to the bottom of the divider. The upper resistor drops $6-4=2\,\mathrm{V}$; the lower drops $4\,\mathrm{V}$. Their drops add to the source voltage.

Read the schematic and results

• Vout — green: midpoint voltage relative to ground; expected 4 V.
• Vin — red: source voltage relative to ground; expected 6 V.
• I — blue: series current, flowing right along the top wire and downward through R1 and R2; expected 2 mA.

The ground symbol at the source return defines $0\,\mathrm{V}$. The voltage probes draw no current in this ideal model. With no branch load, the same current passes through both resistors:

$$\begin{aligned} I &= \frac{V_{\mathrm{in}}}{R_1+R_2}\\ &= \frac{6\,\mathrm{V}}{3\,\mathrm{k}\Omega} =2\,\mathrm{mA} \end{aligned}$$

Ohm's law gives the same output: $V_{\mathrm{out}}=IR_2=4\,\mathrm{V}$. Although both resistors carry the same current, the larger resistance has the larger voltage drop.

The ratio is $V_{\mathrm{out}}/V_{\mathrm{in}}=2/3$. Scaling both resistors by the same factor preserves this unloaded ratio but changes the current. Power is $P=I^2R$: R1 dissipates 4 mW, and R2 dissipates 8 mW. These power values are calculated, not additional table rows.

Use CircuitSim

This example automatically runs DC operating point, which finds the steady voltage and current. Table shows the three probe results in the order Vout, Vin, I. Chart is hidden because a single DC operating point has no changing time or frequency axis.

Select a resistor and change Resistance in Properties, or select V1 and change DCMag (DC voltage). Then use Run simulation to refresh the results. Resistance is in ohms: 1k means $1000\,\Omega$, and 2k means $2000\,\Omega$. Source voltage is in volts.

Try it

1. Make the resistors equal. Change R2's Resistance from 2k to 1k, leaving V1 at 6. Predict the result before running: Vout becomes 3 V, exactly half the input, and I becomes 3 mA.
2. Scale the input. Restore R2 to 2k, then change V1's DCMag (DC voltage) from 6 to 9. Vout becomes 6 V, while I becomes 3 mA. The voltage ratio stays at $2/3$.

Restore V1 = 6, R1 = 1k, and R2 = 2k, then rerun.

Loading changes the ratio

The simple divider formula assumes that no external load draws current from the midpoint. A load connected from Vout to ground is in parallel with R2, lowering the effective bottom resistance and therefore Vout.

For example, adding a $2\,\mathrm{k}\Omega$ load across the existing $2\,\mathrm{k}\Omega$ R2 would make their equivalent resistance $1\,\mathrm{k}\Omega$. With the original $6\,\mathrm{V}$ source and $1\,\mathrm{k}\Omega$ R1, Vout would fall to 3 V. That load is not present in the saved schematic.

A divider sets a voltage ratio; its output also depends on whatever is connected to it. Use the unloaded result when the measuring input draws negligible current.

Model assumptions

The source is ideal and the resistances are constant. Resistor tolerance, temperature effects, source resistance, and measuring-device loading are not included. Keep advanced component parameters at their defaults for these comparisons.