Question

A DC voltage with ripple is given by $v\left(t\right)=[100+10sin(\omega t\left)\right)]volts.$ Measurements of this voltage $v\left(t\right)$, made by moving-coil and moving-iron voltmeters, show readings of $V_1$ and $V_2$ respectively. The value of $V_2-V_1$, in volts, is

• A. 0.312

Answer 1

The correct option is A 0.312
$V\left(t\right)=100+10sin\left(\omega t\right)-5sin\left(3\omega t\right)volt$

Moving coil,
$V_1=V_{avg.}=100V$

Moving iron,
$V_2=V_{rms}=\sqrt{{100}^2+\frac{1}{2}({10}^2+5^2)}$
$=100.312V$

$V_2-V_1=0.312V$