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Q83PE

Expert-verifiedFound in: Page 736

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Find the time after \(t = 0\) when the instantaneous voltage of \(60 - Hz\) \({\rm{AC}}\) first reaches the following values: **

**(a) \({V_0}/2\)**

**(b) \({V_0}\)**

**(c) \(0\)**

- When the instantaneous voltage of \(60\;Hz\)\({\rm{AC}}\) first reaches \(\frac{{{V_0}}}{2}\), the time is \(t = 1.39\;ms\).
- When the instantaneous voltage of \(60\;Hz\)\({\rm{AC}}\) first reaches \({V_0}\), the time is \(t = 4.17\;ms\).
- When the instantaneous voltage of \(60\;Hz\)\({\rm{AC}}\) first reaches \(0\), the time is \(t = 8.33\;ms\).

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

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

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

- The 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 \((1)\)as,

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

Substitute \(\frac{{{V_0}}}{2}\) for \(V\)as,

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

Divide both sides by \({V_0}\) as,

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

Take the inverse sine for both sides as,

\({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = 2\pi ft\)

Divide both sides by \(2\pi f\) as,

\(t = \frac{{{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)}}{{2\pi f}}\)

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

\(\begin{align}{}t &= \frac{{{{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)}}{{2\pi (60\;Hz)}}\\ &= 1.39 \times {10^{ - 3}}\;s\\ &= \left( {1.39 \times {{10}^{ - 3}}\;s} \right)\left( {\frac{{1000\;ms}}{{1\;s}}} \right)\\ &= 1.39\;ms\end{align}\)

Therefore, the value for the time is obtained as \(t = 1.39\;ms\).

b.

Substitute \({V_0}\) for \({\rm{V}}\) into Equation \((1)\) to find the time at which the instantaneous voltage reaches \({V_0}\) as,

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

Divide both sides by \({V_0}\) as,

\(1 = \sin (2\pi ft)\)

Take the inverse sine for both sides as,

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

Divide both sides by \(2\pi f\) as,

\(t = \frac{{{{\sin }^{ - 1}}1}}{{2\pi f}}\)

Entering the values for \({\rm{f}}\), it is obtained as,

\(\begin{align}{}t &= \frac{{{{\sin }^{ - 1}}(1)}}{{2\pi \left( {60{\rm{ }}Hz} \right)}}\\& = 4.17 \times {10^{ - 3}}\;s\\ &= \left( {4.17 \times {{10}^{ - 3}}\;s} \right)\left( {\frac{{1000\;ms}}{{1\;s}}} \right)\\ &= 4.17\;ms\end{align}\)

Therefore, the value for the time is obtained as \(t = 4.17\;ms\).

c.

Substitute zero for \(V\) into Equation \((1)\) to find the time at which the instantaneous voltage reaches zero as,

\(0 = {V_0}\sin (2\pi ft)\)

\(\sin (2\pi ft) = 0\)

The sine has the value of zero when the argument is \(0,\pi ,2\pi , \ldots \). The first time after \(t = 0\) is when \(2\pi ft = \pi \), solving for \(t\) as,

\(\begin{align}{}t &= \frac{1}{{2f}}\\ &= \frac{1}{{2(60\;Hz)}}\\ &= 8.33 \times {10^{ - 3}}\;s\\ &= \left( {8.33 \times {{10}^{ - 3}}\;s} \right)\left( {\frac{{1000\;ms}}{{1\;s}}} \right)\\& = 8.33\;ms\end{align}\)

Therefore, the value for the time is obtained as \(t = 8.33\;ms\).

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