In the circuit shown in the figure, the wiper of the potentiometer performs harmonic oscillatory motion between the two extreme positions. What is the effective value of the voltage measurable between points $A$ and $B$? The internal resistance of the power supply can be neglected. Data: $U=100 \mathrm{~V}, R=R_{1}=1 \mathrm{k} \Omega$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7d438c2feb7b2a473ef2g-1.jpg?height=418&width=585&top_left_y=336&top_left_x=759)
The instantaneous value of the measurable voltage between points $A$ and $B$ is composed of the $U / 2$ voltage dropping across the $R_{1}$ resistor and the $(U / 4) + (U / 4) \sin \omega t$ voltage taken from the $R$ resistor, where $\omega$ is the angular frequency of the harmonic oscillation. Therefore, the voltage measurable between $A$ and $B$ as a function of time is:

$$
U_{p}=(U / 4)(3+\sin \omega t)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f69be7555ee982aeaeb7g-1.jpg?height=236&width=325&top_left_y=435&top_left_x=889)

The definition of the effective voltage:

$$
U_{\text {eff }}^{2}=\frac{1}{T} \int_{0}^{T} U_{p}^{2} \mathrm{~d} t
$$

where $T$ is the period.

Substituting $U_{p}$ and performing the integration using the substitution $\sin ^{2} x=\frac{1-\cos 2 x}{2}$, we get $U_{\text {eff }}=\sqrt{19 / 32} U=$ $77.06 \mathrm{~V}$. György Molnár (Pannonhalma, Bencés Gym., IV. grade)

Note. The value of the effective voltage is independent of the specific values of $R$ and $R_{1}$, provided that $R=R_{1}$.