Question

Suppose the "vertical attenuator" of an A.C signal is set to 0.5v/Div and the probe factor is 10,if signal occupies 5 divisions vertically.What is the peak to peak voltages and the rms equivalent.

Answer 1

Peak to peak voltage equals

$V_{p-p}=K_a\cdot K_p\cdot Y=0.5\cdot 10\cdot 5= 25\space V,$

where $K_a$ - the vertical attenuator factor;

$K_p$ - the probe factor;

Y - number of divisions, occupied by the sygnal.

RMS voltage equals

$V_{rms}=\frac{V_{p-p}}{\sqrt{2}}=\frac{25}{\sqrt{2}}\approx17.7\space V.$

Answer: peak to peak voltage equals 25 V, RMS voltage equals 17.7 V.