Question

In the diagram, $f=50$ Hz. The load voltage is $230$ V (rms) and the load impedance is $\frac{230}{\sqrt{2}}+j\frac{230}{\sqrt{2}}~\Omega$. The attenuator has $A_1=\dfrac{1}{50\sqrt{2}}$. The multiplier output is $v_o=\dfrac{v_x v_y}{1~\text{V}}$. Find the {magnitude of the average value} of $v_o$ (rounded to one decimal place).

Hint:

For sinusoids $v_1=\sqrt{2}V_1\cos(\omega t+\phi_1)$ and $v_2=\sqrt{2}V_2\cos(\omega t+\phi_2)$, the average of their product is $V_1V_2\cos(\phi_1-\phi_2)$.

Answer 1

Correct Answer: 2.1

Load impedance magnitude $|Z|=\sqrt{\left(\frac{230}{\sqrt{2}}\right)^2+\left(\frac{230}{\sqrt{2}}\right)^2}=230~\Omega$.
Hence load (and line) current: $I_{\text{rms}}=\dfrac{230}{230}=1$ A. The $1~\Omega$ series resistor gives \[ V_x(\text{rms})= I_{\text{rms}}\times 1~\Omega=1~\text{V}. \] Because $Z$ has angle $+45^\circ$, current lags $V_{\text{load}}$ by $45^\circ$. The $+90^\circ$ phase shifter makes \[ v_y(\text{rms})=\frac{230}{50\sqrt{2}}=3.2527~\text{V}, \qquad \angle(v_y)-\angle(i) = 90^\circ-(-45^\circ)=135^\circ . \] For two sinusoids at the same frequency, the average of the product equals \[ \overline{v_o}=\frac{V_x V_y}{1~\text{V}}\cos(\Delta\phi) =1\times 3.2527\times \cos 135^\circ =3.2527\times(-0.7071)\approx -2.30~\text{V}. \] Requested magnitude: $\boxed{2.3~\text{V}}$ (to one decimal place).