91Ó°ÊÓ

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.

a.

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

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

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

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\((1)\)–

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

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

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 \(t = 1.04 \times {10^{ - 2}}\;s,1.46 \times {10^{ - 2}}\;s\).