Новое решение
en/11.4.13.md
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| + | ### Statement | ||
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| + | $11.4.13.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + |  | ||
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| + | Let the current through the load (and the resistor $R$) be $I=(12\mbox{ A})e^{i\omega t}$, where $\omega$ is the angular frequency of the source. Then, the voltage drops across the load and the resistor are $V_L=(120\mbox{ V})e^{i(\omega t-\pi/3)}$ and | ||
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| + | \[V_R=IR=(120\mbox{ V})e^{i\omega t},\] | ||
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| + | respectively. Consequently, the EMF of the source is $\mathcal{E}=V_L+V_R$, which becomes | ||
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| + | \[\mathcal{E}=(120\mbox{ V})(e^{i\pi/6}+e^{-i\pi/6})e^{i(\omega t-\pi/6)}=(120\mbox{ V})\cdot2\cos\frac{\pi}{6}\cdot e^{i(\omega t-\pi/6).\] | ||
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| + | Thus, the amplitude of the $EMF$ is $120\sqrt3\approx208$ V. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $11.4.13.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| Let the current through the load (and the resistor $R$) be $I=(12\mbox{ A})e^{i\omega t}$, where $\omega$ is the angular frequency of the source. Then, the voltage drops across the load and the resistor are $V_L=(120\mbox{ V})e^{i(\omega t-\pi/3)}$ and | |||
| \[V_R=IR=(120\mbox{ V})e^{i\omega t},\] | |||
| respectively. Consequently, the EMF of the source is $\mathcal{E}=V_L+V_R$, which becomes | |||
| \[\mathcal{E}=(120\mbox{ V})(e^{i\pi/6}+e^{-i\pi/6})e^{i(\omega t-\pi/6)}=(120\mbox{ V})\cdot2\cos\frac{\pi}{6}\cdot e^{i(\omega t-\pi/6).\] | |||
| Thus, the amplitude of the $EMF$ is $120\sqrt3\approx208$ V. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||