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leonid [27]
3 years ago
14

Lc circuit consists of a 20.0 mh inductor and a 0.150 µf capacitor. if the maximum instantaneous current is 0.400 a, what is the

greatest potential difference across the capacitor?
Physics
1 answer:
slavikrds [6]3 years ago
5 0
The following expression is applicable:

Max. inductor energy = Max. capacitor energy

Where;
Max. inductor energy = LI^2/2, with L = 20.0 mH, I = 0.400 A
Max. capacitor energy = CV_max^2/2, C = 0.150 micro Faraday, V_max = Max. potential difference

Substituting;
LI^2/2 = CV^2/2
LI^2 = CV^2
V^2 = (LI^2)/C
V_max = Sqrt [(LI^2)/C] = Sqrt [(20*10^-3*0.4^2)/(0.15*10^-6)] = 146.06 V
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Answer:

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Explanation:

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L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

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3 years ago
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3 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

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3 years ago
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