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Harlamova29_29 [7]
3 years ago
9

A 20.0-µF capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280

mH inductor. Calculate (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time t = 0 (the time of connection to the inductor); (c) the maximum magnitude of current in the inductor.
Physics
1 answer:
natali 33 [55]3 years ago
7 0

Answer

given,

Capacitance of capacitor = 20.0-µF

Voltage =  150.0-V

inductance = 0.280 m H

a) the oscillation frequency of circuit

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

f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{0.280 \times 10^{-3}\times 20 \times 10^{-6}}}

f = 2126.9 Hz

b) U = \dfrac{1}{2}CV^2

   U = \dfrac{1}{2}(20\times 10^{-6})(150)^2

         U = 0.225 J

c)Current in the inductor

    V = L \dfrac{dI}{dt}

    150 = 0.28 \times 10^{-3} \dfrac{dI}{dt}

    \dfrac{dI}{dt} = 535714

instantaneous rate of change of current is equal to 535714 A/s

 

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