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Q84PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) At what two times in the first period following \(t = 0\) does the instantaneous voltage in \(60 - Hz\) \({\rm{AC}}\) equal \({V_{rms}}\) ? **

**(b) \({V_{rms}}\)?**

- At times \(t = 2.08 \times {10^{ - 3}}\;s,6.25 \times {10^{ - 3}}\;s\) in the first period following \(t = 0\) , the instantaneous voltage in \(60{\rm{ }}Hz\)\({\rm{AC}}\) equal \({V_{rms}}\).
- At times \(t = 1.04 \times {10^{ - 2}}\;s,1.46 \times {10^{ - 2}}\;s\) in the first period following \(t = 0\) , the instantaneous voltage in \(60{\rm{ }}Hz\)\({\rm{AC}}\) equal \( - {V_{rms}}\).

**The potential difference between the terminals of an **\({\rm{AC}}\)** voltage source is expressed mathematically as a function of time as follows as,**

\(V(t) = {V_0}\sin (2\pi ft)...(1)\)

**Where **\(V(t)\)** is the voltage at the time **\(t\)**, **\({V_0}\)** is the peak voltage, and **\(f\)** is the frequency in Hertz.**

**The rms voltage in an **\({\rm{AC}}\)** circuit in which the voltage varies sinusoidally is given as,**

\(\begin{align}{}{V_{rms}} = \frac{{{V_0}}}{{\sqrt 2 }}\\ = 0.707{V_0}...(2)\end{align}\)

**Where **\({V_0}\)** is the maximum voltage.**

- The frequency of the \({\rm{AC}}\) source is: \(f = 60{\rm{ }}Hz\).

a.

The voltage of the \({\rm{AC}}\) source is expressed as a function of time from Equation \((1)\)–

\(V = {V_0}\sin (2\pi ft)\)

Substitute \({V_{rms}}\) for \(V\) –

\({V_{rms}} = {V_0}\sin (2\pi ft)\)

Where \({V_{rms}}\) is found from Equation \((2)\)–

\(\frac{{{V_0}}}{{\sqrt 2 }} = {V_0}\sin (2\pi ft)\)

Divide both sides by \({V_0}\) –

\(\frac{1}{{\sqrt 2 }} = \sin (2\pi ft)\)

Take the inverse sine for both sides –

\(\begin{align}{}{\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = 2\pi ft\\2\pi ft = \frac{\pi }{4},\frac{{3\pi }}{4}\end{align}\)

Divide both sides by \(2\pi f\) –

\(t = \frac{1}{{8f}},\frac{3}{{8f}}\)

Entering the values for \(f\), it is obtained –

\(\begin{align}{}t = \frac{1}{{8(60\;Hz)}},\frac{3}{{8(60\;Hz)}}\\ = 2.08 \times {10^{ - 3}}\;s,6.25 \times {10^{ - 3}}\;s\end{align}\)

Therefore, the values for the time is obtained as \(t = 2.08 \times {10^{ - 3}}\;s,6.25 \times {10^{ - 3}}\;s\).

b.

The voltage of the \({\rm{AC}}\) source is expressed as a function of time from Equation \((1)\)–

\({V_{rms}} = {V_0}\sin (2\pi ft)\)

Substitute \({V_{rms}}\) for \(V\) –

\( - {V_{rms}} = {V_0}\sin (2\pi ft)\)

Where \({{\rm{V}}_{{\rm{rms}}}}\) is found from Equation \((2)\)–

\( - \frac{{{V_0}}}{{\sqrt 2 }} = {V_0}\sin (2\pi ft)\)

Divide both sides by \({V_0}\) –

\( - \frac{1}{{\sqrt 2 }} = \sin (2\pi ft)\)

Take the inverse sine for both sides –

\(\begin{align}{}{\sin ^{ - 1}}\left( { - \frac{1}{{\sqrt 2 }}} \right) = 2\pi ft\\2\pi ft = \frac{{5\pi }}{4},\frac{{7\pi }}{4}\end{align}\)

Divide both sides by \(2\pi f\) –

\(t = \frac{5}{{8f}},\frac{7}{{8f}}\)

Entering the values for \(f\), it is obtained –

\(\begin{align}{}t = \frac{5}{{8(60\;Hz)}},\frac{7}{{8(60\;Hz)}}\\ = 1.04 \times {10^{ - 2}}\;s,1.46 \times {10^{ - 2}}\;s\end{align}\)

Therefore, the values for the time is obtained as \(t = 1.04 \times {10^{ - 2}}\;s,1.46 \times {10^{ - 2}}\;s\).

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