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Can anyone solve this? I've asked 3 professors

  • Thread starter Moore65
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Can anyone solve this? I've asked 3 professors!!!

I'm having trouble trying to understand this problem.

The R-L-C series circuit is driven with a AC source with EMF=Vo*sin(wt) where Vo=110.0V and f=60.0 Hz. If R = 20.0 Ohms, L=5.0 * 10^-2 H, and C=50.0uF, find the potential drops across the inductor at to=0 and at a time t1=the first time after to that the EMF reaches a maximum.

The current moves throught the circuit first through the resistor, then through the capacitor, and finally through the "L" before completely the cicuit.

I'm not really sure what to do, can anyone help?
 

Answers and Replies

  • #2
Tide
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The sum of the potential drops across the resistor, capacitor and inductor equals the applied voltage and can be expressed as a differential equation. Just solve it for the current and find the potential drop across the inductor as L dI/dt.
 
  • #3
ehild
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Moore65 said:
I'm having trouble trying to understand this problem.

The R-L-C series circuit is driven with a AC source with EMF=Vo*sin(wt) where Vo=110.0V and f=60.0 Hz. If R = 20.0 Ohms, L=5.0 * 10^-2 H, and C=50.0uF, find the potential drops across the inductor at to=0 and at a time t1=the first time after to that the EMF reaches a maximum.

The current moves throught the circuit first through the resistor, then through the capacitor, and finally through the "L" before completely the cicuit.

I'm not really sure what to do, can anyone help?

What have you learnt about ac circuits? Do you know what "impedance" is?

The magnitude of the impedance of a series L-R-C circuit is

[tex] Z=\sqrt{R^2+(\omega L - 1/(\omega C))^2} [/tex]

and the phase of the impedance is

[tex]\alpha = \arctan ( \frac{\omega L - 1/(\omega C)}{R})[/tex]

The amplitude of the current through the circuit is

[tex] I_0 = U_0 /Z [/tex]

The phase of the impedance is equal to the phase difference between voltage and current: the current lags behind the voltage by this phase angle. The time dependence of the current is:

[tex] I(t) = I_0 \sin(\omega t - \alpha) [/tex]

The potential across the inductor leeds the current by pi/2. The impedance of the inductor is wL.

[tex] U_L = (\omega L) I_0 \sin (\omega t -\alpha + \ pi /2)[/tex]

First calculate Z, alpha, Io and then plug in t=0 to get the potential drop through the inductor at t=0. The emf reaches its maximum first when wt= pi/2, so just replace (wt) by pi/2 to get the potential drop at t1. .

ehild
 

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