Suggested languages for you:

Americas

Europe

Q83PE

Expert-verified
Found in: Page 736

College Physics (Urone)

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) $$0$$

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

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

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

Step 2: Information Provided

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

Step 3: Calculation for time ($${V_0}/2$$)

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 $$t = 1.39\;ms$$.

Step 4: Calculation for time ($${V_0}$$)

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 $$t = 4.17\;ms$$.

Step 5: Calculation for time ($$0$$)

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 $$t = 8.33\;ms$$.