Question

The waveform of a voltage in an electrical circuit has the following parameters:
 A continuous series of right-angled triangles, each with a base length equivalent to 3 ms
 9 V maximum
 Voltage ramps up over the 3 ms and then drops back to zero instantaneously
 Sketch the waveform and mark on the relevant parameters.
 Derive the function which defines the waveform.
 Using integral calculus, find the RMS value of the voltage.

Answer 1

Answer on Question # 76724, Physics -Electric Circuits:

Question: The waveform of a voltage in an electrical circuit has the following parameters:

i. A continuous series of right-angled triangles, each with a base length equivalent to 3 ms

ii. $9\mathrm{V}$ maximum

iii. Voltage ramps up over the 3 ms and then drops back to zero instantaneously

iv. Sketch the waveform and mark on the relevant parameters.

V. Derive the function which defines the waveform.

Vi. Using integral calculus, find the RMS value of the voltage.

Solution: Wave function is given by :

In the above wave function green line indicates the time and red line indicates the maximum amplitude of the signal.

Here, maximum amplitude $V_{p} = 9$ volt.

Time period $(T) = 3\mathrm{ms}$

Wave function is given by:

$$\begin{array}{l} U = \frac{9 - 0}{3 - 0} t \\ = 3 t \quad 0 \leq t < 3 \end{array}$$

Now, let $r$ be the voltage and $U_{r}$ is the value of $U$.

$$U_{r} = \frac{1}{T} \int [9t^2] \, dt$$

where $U_{r}$ is the voltage and $U_{r}$ is the value of $U$. The integral limit is 0 to 3 and $T = 3$.

Now, $U_{r}^{2} = 9$ or $U_{r}^{2} = 3$ volts.

Answer: $r$ is the voltage and $U_{r}$ is the value of $U$.