# Effective and mean voltage

A voltage divider consisting of resistors R1 = 103000 Ω and R2 = 197000 Ω is connected to the ideal sine wave voltage source, R2 is connected to a voltmeter which measures the mean voltage and has an internal resistance R3 = 200300 Ω, the measured value is 109 V. Determine the effective source voltage.

Result

U1 =  246.625 V

#### Solution:

$R_{1}=103000 \ Ω \ \\ R_{2}=197000 \ Ω \ \\ R_{3}=200300 \ Ω \ \\ \ \\ U_{2}=109 \ \text{V} \ \\ U_{21}=\pi \cdot \ U_{2}/2=3.1416 \cdot \ 109/2 \doteq 171.2168 \ \text{V} \ \\ U_{22}=U_{21}/\sqrt{ 2 }=171.2168/\sqrt{ 2 } \doteq 121.0686 \ \text{V} \ \\ \ \\ I_{3}=U_{22}/R_{3}=121.0686/200300 \doteq 0.0006 \ \text{A} \ \\ I_{2}=U_{22}/R_{2}=121.0686/197000 \doteq 0.0006 \ \text{A} \ \\ I_{1}=I_{2}+I_{3}=0.0006+0.0006 \doteq 0.0012 \ \text{A} \ \\ \ \\ U_{11}=I_{1} \cdot \ R_{1}=0.0012 \cdot \ 103000 \doteq 125.5567 \ \text{V} \ \\ \ \\ U_{1}=U_{11} + U_{22}=125.5567 + 121.0686 \doteq 246.6253 \doteq 246.625 \ \text{V}$

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