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rjkz [21]
3 years ago
9

Calculate the reactance of a 0.5 F capacitor that is connected to a battery with peak voltage 2 V and angular frequency 200 radi

ans/s.
Physics
1 answer:
Furkat [3]3 years ago
6 0

Answer:

Reactance of the capacitor is 0.01 ohms.

Explanation:

It is given that,

Capacitance of the capacitor is 0.5 F

Peak voltage of the battery is 2 V

Angular frequency is 200 rad/s

We need to find the capacitive reactance. The capacitive reactance is due to the capacitor in the circuit. It is given by :

X_c=\dfrac{1}{\omega C}\\\\X_c=\dfrac{1}{200\times 0.5}\\\\X_c=0.01\ \Omega

So, the reactance is 0.01 ohms.

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The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?
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Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, L=9.4\ mH=9.4\times 10^{-3}\ H

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Capacitance, C=1.9\ \mu F=1.9\times 10^{-6}\ C

At resonance, the capacitive reactance is equal to the inductive reactance such that,

X_C=X_L    

2\pi f_o L=\dfrac{1}{2\pi f_oC}

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f_o=\dfrac{1}{2\pi \sqrt{LC}}

f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}

f_o=1190.91\ Hz

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

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