Question

A sine wave with a rms value of 12V is riding on dc level of 18V. The maximum value of resulting waveform is
(A). 6V
(B). 30V
(C). 35V
(D). 0V

Answer 1

Hint: An electric current which changes its magnitude continuously and direction periodically with respect to time is called alternating current. And the voltage which changes its magnitude continuously and direction periodically with respect to time is called alternating voltage.
On the other hand, if an electric current is unidirectional means electric charges flowing only in one direction with constant magnitude are called direct current.

Complete step by step answer:
Consider an alternating current its instantaneous voltage is given by
$V={{V}_{0}}\sin (\omega t)$
Here, V is instantons voltage
${{V}_{0}}$ is the peak voltage (maximum voltage)
$\omega$ and t are angular frequency and time, respectively.
Now we have to find ${{V}_{0}}$by using ${{V}_{rms}}$ whose value is given in the question
Even though, maximum value of an AC is ${{V}_{0}}$ since it is riding on a DC voltage , its maximum value is given by:
${{V}_{\max }}={{V}_{0}}+{{V}^{\iota }}$ ……. (1)
Where ${{V}^{\iota }}$is the DC level value.
To solve for ${{V}_{0}}$, consider RMS value,
RMS value: It is the root mean square of a sinusoidal wave which is equal to that direct current which produces the same heating in resistance as is produced by A.C. in same resistance during same time. It is called effective value.
${{V}_{0}}$ and ${{V}_{rms}}$ are related by the formula,
${{V}_{rms}}=\dfrac{{{V}_{0}}}{\sqrt{2}}$ ….. (2)
Where, ${{V}_{m}}$ is the peak value of the alternating current.
$\Rightarrow {{V}_{0}}={{V}_{rms}}\sqrt{2}$
Now substituting ${{V}_{rms}}$=12 V, we get
$\begin{aligned} & {{V}_{0}}=12\times \sqrt{2}=12\times 1.414 \\\ & {{V}_{0}}=16.96 \\\ \end{aligned}$
In the given question it is given that ${{V}^{\iota }}$=18V.
substituting these two values in (1), we get
$\begin{aligned} & {{V}_{0}}=16.95+18 \\\ & {{V}_{0}}=34.95\approx 35V \\\ \end{aligned}$
Thus, the maximum value of the resulting waveform is =35V and the correct option is C.

Note: Similarly, minimum value of an AC is given by,
${{V}_{\min }}={{V}_{0}}-{{V}^{\iota }}$ where ${{V}_{0}}$and ${{V}^{\iota }}$ are the peak value and DC level value respectively.
Students may get confused between ${{V}_{rms}}$ and ${{V}_{0}}$ while writing formula, rms value is effective value and related to peak value as in equation(2).
Totally we have four types of voltage in AC:
Instantaneous voltage.
Peak voltage
RMS voltage- which are all discussed above and one more is,
Average voltage -which is given by,
For complete cycle – 0
For half cycle, ${{V}_{avg}}=\dfrac{2{{V}_{0}}}{\pi }$.