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Expert-verified Found in: Page 736 ### College Physics (Urone)

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) $${V_{rms}}$$?

1. 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 $${V_{rms}}$$.
2. 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 $$- {V_{rms}}$$.
See the step by step solution

## Step 1: Concept Introduction

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.

## Step 2: Information Provided

• The frequency of the $${\rm{AC}}$$ source is: $$f = 60{\rm{ }}Hz$$.

## Step 3: Calculation for time ($${{\rm{V}}_{{\rm{rms}}}}$$)

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 $$t = 2.08 \times {10^{ - 3}}\;s,6.25 \times {10^{ - 3}}\;s$$.

## Step 4: Calculation for time ($${\rm{ - }}{{\rm{V}}_{{\rm{rms}}}}$$)

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 $$t = 1.04 \times {10^{ - 2}}\;s,1.46 \times {10^{ - 2}}\;s$$. ### Want to see more solutions like these? 