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| 184_notes:examples:week14_ac_graph [2017/11/28 16:50] – [Solution] tallpaul | 184_notes:examples:week14_ac_graph [2017/11/28 16:51] – tallpaul | ||
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| The period of the graph is given by the time between peaks. One can see that the graph can be split into vertical slices as shown below. | The period of the graph is given by the time between peaks. One can see that the graph can be split into vertical slices as shown below. | ||
| - | {{pic}} | + | {{ 184_notes: |
| It should be easy to see based on the visual above that the time between peaks is just four times the time value at the first peak. So then the period is $0.04\text{ s}$. | It should be easy to see based on the visual above that the time between peaks is just four times the time value at the first peak. So then the period is $0.04\text{ s}$. | ||
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| $$f = \frac{1}{\tau} = \frac{1}{0.04\text{ s}} = 25\text{ Hz}$$ | $$f = \frac{1}{\tau} = \frac{1}{0.04\text{ s}} = 25\text{ Hz}$$ | ||
| - | We are also now equipped to write a equation for the alternating current. We will use a sine function rather than a cosine function, since the current begins at $I=0$, and increasing. The amplitude will provide the factor out front: | + | We are also now equipped to write a equation for the alternating current. We will use a sine function rather than a cosine function, since the current begins at $I=0$, and increases. The amplitude will provide the factor out front: |
| $$I=I_0\sin(2\pi \cdot f \cdot t) = (0.3\text{ A}) \cdot \sin(2\pi \cdot 25\text{ Hz} \cdot t)$$ | $$I=I_0\sin(2\pi \cdot f \cdot t) = (0.3\text{ A}) \cdot \sin(2\pi \cdot 25\text{ Hz} \cdot t)$$ | ||