# The output of a certain voltage divider is 12 V with no load. When a load is connected, does the output voltage decrease?

Jul 26, 2018

Yes

#### Explanation:

The voltage at the output of the voltage divider is determined by the voltage dropped across the resistors in the divider.

With no load, the current flowing in ${R}_{1}$ is

I_(R_1) = V_("in")/(R_1+R_2) " "(= I_(R_2))

If a load (${R}_{L}$) is connected to the output, (across ${R}_{2}$) the resistance at the output decreases from ${R}_{2}$, to ${R}_{2}$ in parallel with ${R}_{L}$.

So I_(R_(1_L)) = V_("in")/(R_1+(R_2|\|R_L)

$\left({R}_{2} | \setminus | {R}_{L}\right) < {R}_{2} \text{, so } {I}_{{R}_{{1}_{L}}} > {I}_{{R}_{1}}$

So we see that the current through ${R}_{1}$ increases when a load is connected.

The voltage dropped across ${R}_{1}$ therefore increases, so the voltage at the output decreases.