- #1
Lunat1c
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Consider the circuit show below, where V1 = V_{peak}sin(\omega t - 36^\circ)
http://img195.imageshack.us/img195/9687/rlc.png
I wish to find an expression for I in the time domain and in phasor form and then sketch the magnitude/phase vs frequency plots. I already managed to get a solution but it seems way too complex to be able to sketch the above mentioned plots from it. So I'm just going to list my solution and perhaps someone will be kind enough to suggest other ways to tackle this problem (if there are any).
to find i(t):
V1(t) = i(t)R_1 + i(t)\bigg[2 \pi fL || \frac{1}{2 \pi fC} || R_2\bigg]
But f = \frac{\omega}{2 \pi}
Hence, V1(t) = i(t)R_1 + i(t)\bigg[\omega L || \frac{1}{\omega C} || R_2\bigg] = i(t)\bigg[R_1 + \frac{\omega LR_2}{R_2 + \omega^2 LCR_2 + \omega L}\bigg]
After substituting for V1 and simplifying, I ended up with:
i(t) = \frac{\omega^2 LCR_2 + \omega L + R_2}{\omega^2 LCR_1R_2 + \omega L(R_1+R_2) + R_1R_2}V_{peak} sin(\omega t - 36^\circ)
Which seems ok.
Now, for the phasor part:
v(j \omega) = i(j \omega)R_1 + i(j \omega)\bigg[j\omega L || \frac{1}{j\omega C} || R_2\bigg] = i(j \omega)R_1 + i(j\omega)\bigg[\frac{1}{\frac{1}{j \omega L} + j \omega C + \frac{1}{R_2}}\bigg] = i(j\omega)\bigg(R_1 + \frac{j\omega LR_2}{R_2 - \omega^2 LCR_2 + j\omega L}\bigg) = i(j \omega)\bigg(\frac{R_1R_2 + j \omega LR_1 - \omega^2 LCR_1R_2 + j\omega LR_2}{R_2 + j \omega L - \omega^2 LCR_2}\bigg)
Hence, i(j \omega) = \frac{R_2(1 - \omega^2 LC) + j \omega L}{R_1R_2 + j \omega L (R_1+R_2) - \omega^2 LCR_1R_2}v(j \omega)
And now I proceded to finding |\bar{I}| and \phi
Note that V1 = V_{peak}sin(\omega t - 36^\circ)= V_{peak} \angle{-36^\circ}
\bar{I} = \frac{\sqrt{R_2^2(1 - \omega^2 LC)^2 + (\omega L)^2}}{\sqrt{R_1^2R_2^2(1 - LC \omega^2)^2 + (\omega L)^2(R_1 + R_2)^2}} V_{peak}
and
\phi = tan^{-1}\frac{\omega L}{R_2(1 - \omega^2 LC)} - 36^\circ - tan^{-1}\frac{\omega L(R_1+R_2)}{R_1R_2 - \omega^2 LCR_1R_2}
How would one go about sketching this without the help of any software package? There must be another way to solve this problem which is way simpler. Any ideas? Any help will be greatly appreciated
http://img195.imageshack.us/img195/9687/rlc.png
I wish to find an expression for I in the time domain and in phasor form and then sketch the magnitude/phase vs frequency plots. I already managed to get a solution but it seems way too complex to be able to sketch the above mentioned plots from it. So I'm just going to list my solution and perhaps someone will be kind enough to suggest other ways to tackle this problem (if there are any).
to find i(t):
V1(t) = i(t)R_1 + i(t)\bigg[2 \pi fL || \frac{1}{2 \pi fC} || R_2\bigg]
But f = \frac{\omega}{2 \pi}
Hence, V1(t) = i(t)R_1 + i(t)\bigg[\omega L || \frac{1}{\omega C} || R_2\bigg] = i(t)\bigg[R_1 + \frac{\omega LR_2}{R_2 + \omega^2 LCR_2 + \omega L}\bigg]
After substituting for V1 and simplifying, I ended up with:
i(t) = \frac{\omega^2 LCR_2 + \omega L + R_2}{\omega^2 LCR_1R_2 + \omega L(R_1+R_2) + R_1R_2}V_{peak} sin(\omega t - 36^\circ)
Which seems ok.
Now, for the phasor part:
v(j \omega) = i(j \omega)R_1 + i(j \omega)\bigg[j\omega L || \frac{1}{j\omega C} || R_2\bigg] = i(j \omega)R_1 + i(j\omega)\bigg[\frac{1}{\frac{1}{j \omega L} + j \omega C + \frac{1}{R_2}}\bigg] = i(j\omega)\bigg(R_1 + \frac{j\omega LR_2}{R_2 - \omega^2 LCR_2 + j\omega L}\bigg) = i(j \omega)\bigg(\frac{R_1R_2 + j \omega LR_1 - \omega^2 LCR_1R_2 + j\omega LR_2}{R_2 + j \omega L - \omega^2 LCR_2}\bigg)
Hence, i(j \omega) = \frac{R_2(1 - \omega^2 LC) + j \omega L}{R_1R_2 + j \omega L (R_1+R_2) - \omega^2 LCR_1R_2}v(j \omega)
And now I proceded to finding |\bar{I}| and \phi
Note that V1 = V_{peak}sin(\omega t - 36^\circ)= V_{peak} \angle{-36^\circ}
\bar{I} = \frac{\sqrt{R_2^2(1 - \omega^2 LC)^2 + (\omega L)^2}}{\sqrt{R_1^2R_2^2(1 - LC \omega^2)^2 + (\omega L)^2(R_1 + R_2)^2}} V_{peak}
and
\phi = tan^{-1}\frac{\omega L}{R_2(1 - \omega^2 LC)} - 36^\circ - tan^{-1}\frac{\omega L(R_1+R_2)}{R_1R_2 - \omega^2 LCR_1R_2}
How would one go about sketching this without the help of any software package? There must be another way to solve this problem which is way simpler. Any ideas? Any help will be greatly appreciated
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