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| 184_notes:examples:week14_ac_graph [2017/11/28 16:50] – [Solution] tallpaul | 184_notes:examples:week14_ac_graph [2018/08/09 19:19] (current) – curdemma |
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| | [[184_notes:ac|Return to Changing Flux from an Alternating Current notes]] |
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| ===== Analyzing an Alternating Current Graph ===== | ===== Analyzing an Alternating Current Graph ===== |
| Suppose you are given the following graph of current over time. You can see that the first peak is at the point where t=0.01 s, and I=0.3 A. The graph is shown below. Find the amplitude, period, and frequency of the current, and give an equation that describes the alternating current. | Suppose you are given the following graph of current over time. You can see that the first peak is at the point where t=0.01 s, and I=0.3 A. The graph is shown below. Find the amplitude, period, and frequency of the current, and give an equation that describes the alternating current. |
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| {{ 184_notes:14_ac_graph_given.png?500 |Graph of Alternating Current}} | [{{ 184_notes:14_ac_graph_given.png?400 |Graph of Alternating Current}}] |
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| ===Facts=== | ===Facts=== |
| 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. |
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| {{pic}} | [{{ 184_notes:14_ac_graph_slices.png?400 |Vertical Time Slices}}] |
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| 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 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 s. |
| f=1τ=10.04 s=25 Hz
| f=1τ=10.04 s=25 Hz
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| 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: |
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| I=I0sin(2π⋅f⋅t)=(0.3 A)⋅sin(2π⋅25 Hz⋅t)
| I=I0sin(2π⋅f⋅t)=(0.3 A)⋅sin(2π⋅25 Hz⋅t)
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