Question

Household electricity is supplied in the form of an alternating current that varies from 155V to -155V with a frequency of cycles per second (Hz). The voltage is thus given by the equation, \(E(t) = 155\sin \left( {120\pi t} \right)\) where t is the time in seconds. Voltmeters read the RMS (root mean square) voltage which is the square root of the average value of \(E{\left( t \right)^2}\) over one cycle.

1. Calculate the RMS voltage of the household current
2. Many electric stoves require an RMS voltage of 220V. Find the corresponding amplitude A needed for the voltage \(E(t) = \sin \left( {120\pi t} \right)\)

Short answer

1. The RMS voltage of household current with the voltage equation \(E(t) = 155\sin \left( {120\pi t} \right)\) is

b. The value of the Amplitude A needed for the voltage \(E(t) = \sin \left( {120\pi t} \right)\)with an RMS voltage of 220V is

Step-by-step solution

Step-1: Calculating average voltage

Given that, the voltage equation is of \(E(t) = 155\sin \left( {120\pi t} \right)\)

Squaring the above equation \({\left( {E(t)} \right)^2} = {\left( {155} \right)^2}{\sin ^2}\left( {120\pi t} \right)\)

Finding the average of the squared equation over one cycle from 0 to 60Hz i.e., 0 to 1/60 seconds of time interval

i.e.,

\(\begin{aligned}{l}{V_{avg}} &= \frac{1}{{\frac{1}{{60}} - 0}}\int\limits_0^{\frac{1}{{60}}} {{{\left( {E\left( t \right)} \right)}^2}dt} \\ &= 60\int\limits_0^{\frac{1}{{60}}} {{{(155)}^2}} {\sin ^2}(120\pi t)dt\\ &= 60{(155)^2}\int\limits_0^{\frac{1}{{60}}} {{{\sin }^2}} (120\pi t)dt\end{aligned}\)

Since we have \({\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\)

\(\begin{aligned}{l} &= 60{(155)^2}\int\limits_0^{\frac{1}{{60}}} {\left( {\frac{{1 - \cos \left( {240\pi t} \right)}}{2}} \right)} dt\\ &= \frac{{1441500}}{2}\int\limits_0^{\frac{1}{{60}}} {\left( {1 - \cos \left( {240\pi t} \right)} \right)} dt\\ &= 720750\left( {t - \frac{{\sin (240\pi t)}}{{240\pi }}} \right)_0^{\frac{1}{{60}}}\end{aligned}\)

Step-2: Simplifying and applying limits

\(\begin{aligned}{l}{V_{avg}} &= 720750\left( {\left( {\frac{1}{{60}} - \frac{{\sin \left( {240\pi \left( {\frac{1}{{60}}} \right)} \right)}}{{240\pi }}} \right) - \left( {0 - \frac{{\sin 0}}{{240\pi }}} \right)} \right)\\{V_{avg}} &= 720750\left( {\left( {\frac{1}{{60}} - \frac{{\sin \left( {4\pi } \right)}}{{240\pi }}} \right) - 0} \right)\\{V_{avg}} &= 720750\left( {\frac{1}{{60}} - 0} \right)\\{V_{avg}} &= 12012.5\end{aligned}\)

Step-3: Calculating the RMS

RMS voltage \( = \sqrt {Averagevoltage} \)

\(\begin{aligned}{l} &= \sqrt {{V_{avg}}} \\ &= \sqrt {12012.5} \\ &= 109.60V\end{aligned}\)

Therefore, the RMS voltage of household current with a voltage equation \(E(t) = 155\sin \left( {120\pi t} \right)\) is

Step-4: Given data for subpart (b)

The RMS voltage required for the electric stoves is of 220V

The voltage equation is \(E(t) = A\sin \left( {120\pi t} \right)\) where A is the amplitude

Step-5: Calculating the amplitude using RMS voltage

We have RMS voltage \( = \frac{{Peakvoltage}}{{\sqrt 2 }}\) for sinusoidal wave

Thus, here peak voltage=A

\(\)

\( \Rightarrow RMS = \frac{A}{{\sqrt 2 }}\)

On substituting the values,

Therefore, the amplitude needed for the voltage \(E(t) = A\sin \left( {120\pi t} \right)\) with an RMS Voltage 220V is