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

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

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Answer 1 · 2018-07-26T03:09:25

Yes

Explanation:

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

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[image source: https://www.allaboutcircuits.com/tools/voltage-divider-calculator/ ]

With no load, the current flowing in $R_{1}$ $R_1$ is

$I_{R_{1}} = \frac{V_{in}}{R_{1} + R_{2}} (= I_{R_{2}})$ $I_(R_1) = V_("in")/(R_1+R_2) " "(= I_(R_2))$

If a load ( $R_{L}$ $R_L$ ) is connected to the output, (across $R_{2}$ $R_2$ ) the resistance at the output decreases from $R_{2}$ $R_2$ , to $R_{2}$ $R_2$ in parallel with $R_{L}$ $R_L$ .

So $I_{R_{1_{L}}} = \frac{V_{in}}{R_{1} + (R_{2} ∣ ∣ R_{L})}$ $I_(R_(1_L)) = V_("in")/(R_1+(R_2|\|R_L)$

$(R_{2} ∣ ∣ R_{L}) < R_{2}, so I_{R_{1_{L}}} > I_{R_{1}}$ $(R_2|\|R_L) < R_2 ", so " I_(R_(1_L)) > I_(R_1)$

So we see that the current through $R_{1}$ $R_1$ increases when a load is connected.

The voltage dropped across $R_{1}$ $R_1$ therefore increases, so the voltage at the output decreases.

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

1 Answer

Explanation:

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