What is the reactance of the capacitor at 1.00 kHz?

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SUMMARY

The reactance of a capacitor at 1.00 kHz can be calculated using the formula Xc = 1/(ωC), where ω is the angular frequency (2πf). Given an rms voltage of 0.550 V and an rms current of 0.460 mA, the reactance is determined to be 5.25 kΩ, leading to a capacitance of 0.0303 µF. For higher frequencies, the rms current can be calculated using Irms = ωCVrms, resulting in 0.208 mA at 2.00 kHz and 1.04 mA at 10.0 kHz.

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i seemto be frequenting this board more than i intended to but i physics is not my strong point at all so if i seem bothersome apologize about that and i also try to put what i have done sor for in the problem so it don;t seem like i am not trying so here goes and besides does anyone know where i can get a darn good physics tutor, one i can talk to over the phone?

A circuit consists of a 1.00 kHz generator and a capacitor. When the rms voltage of the generator is 0.550 V the rms current in the circuit is 0.460 mA.
(a) What is the reactance of the capacitor at 1.00 kHz?
k
(b) What is the capacitance of the capacitor?
µF
(c) If the rms voltage is maintained at 0.550 V, what is the rms current at 2.00 kHz?
mA
If the rms voltage is maintained at 0.550 V, what is the rms current at 10.0 kHz?
mA

1. to get the reactance first i derived c from eqn Irms= Vrms * wc
and then sub for Xc= 1/above ans which was 5.25)*2(3.140*1000hz
that should have given me the ans but the book says different

for the b capacitance, i used Xc= 1/wc

for c it should be irms= wCVIrms - right

thanks
 
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I don't remember much about alternating current but, isn't impedance:
Z = \sqrt{R^2 + X^2} = \frac{V}{I}, where X is the total reactance (X_C + X_L).
As there's only a capacitor (and a generator) in the circuit the above equation solves into Z = \frac{V}{I} = X_C

You don't need X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} for a)
 
for your post!

To find the reactance of the capacitor at 1.00 kHz, you can use the formula Xc = 1/(wC), where w is the angular frequency (2πf) and C is the capacitance. In this case, w = 2π(1.00 kHz) = 6.28 kHz. Using the given values of Irms = 0.460 mA and Vrms = 0.550 V, you can rearrange the formula to solve for C: C = 1/(wXc) = 1/(6.28 kHz * 5.25) = 0.0303 µF.

For part (c), you are correct that the formula Irms = wCVrms can be used. In this case, w = 2π(2.00 kHz) = 12.56 kHz. Plugging in the values of C = 0.0303 µF and Vrms = 0.550 V, we get Irms = (12.56 kHz * 0.0303 µF * 0.550 V) = 0.208 mA.

Similarly, for 10.0 kHz, w = 2π(10.0 kHz) = 62.8 kHz. Plugging in the same values of C and Vrms, we get Irms = (62.8 kHz * 0.0303 µF * 0.550 V) = 1.04 mA.

As for finding a good physics tutor, I would recommend reaching out to your school's tutoring center or academic support services. They may have resources available for phone or online tutoring. You can also try searching for online tutoring services or reaching out to physics departments at nearby universities to see if they offer tutoring services. Best of luck in your studies!
 

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