Question

The instantaneous value of an alternating voltage in volts is given by the expression $E_{t} = 140\ sin 300t$ where t is in second what is (i) peak value of the voltage (ii) Its rms value and (iii) frequency of the supply? Take $\pi = 3$, $\sqrt{2} = 1.4$

Answer 1

Hint: The highest instantaneous value of a function is estimated from the zero-volt level. The peak value of amplitude and peak value are equal since the average value of the function is zero volts. The term "RMS" is "Root-Mean-Squared," also named for the practical value of alternating current, similar to a DC voltage that would give the same amount of heat production in a resistor as the AC voltage would if implemented to that equivalent resistor.

Complete step by step solution:
Given: The instantaneous value of an alternating voltage:
$E_{t} = 140\ sin 300t$
We will compare this equation with the general equation:
$E_{t} = E_{o}\ sin 300t$
Where, $ E_{o} $ is the peak value of voltage.
$ E_{t} $ is the value of voltage at any time t.
w is the angular frequency.
After comparing, we get,
$ E_{o} = 140 V$
$w = 300$
a) Peak value of voltage, $ E_{o} = 140 V$
b) Rms value of voltage,
$E_{rms} = \dfrac{E_{o}}{\sqrt{2}}$
Put the value of $ E_{o} = 140 V$.
$E_{rms} = \dfrac{140}{\sqrt{2}} = \dfrac{140}{1.4}$
$E_{rms} = 100V$
c) Angular frequency, $w = 300$
Now, we will find frequency by the formula:
$f = \dfrac{w}{2 \pi}$
Put $w = 300$
$f = \dfrac{300}{2 \times 3 }$
We get,
$f = 50 Hz$

Note: RMS is not the "Average" voltage, and its numerical relationship to peak voltage varies on the type of waveform. The RMS value is squared of the root of the mean value of the squared function of the instant values. Rms value is the amount of AC power that provides the same heating effect as an equal DC power.