Question

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.

Answer 1

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

 

Answer 2

The value for each optuon is mathematically given as

a) f = 2126.9 Hz

b) U = 0.225 J

c) dI/dt= 535714A/s

What is a power supply?

A power supply is simply defined as an electrical device that gives electric power to an electrical output.

Question Parameter(s):

A 20.0-µF capacitor is charged by a 150.0-V power supply, then

in series with a 0.280 mH

Generally, the equation for the Frequency   is mathematically given as

$f = \frac{1}{2\pi}\sqrt{\frac{1}{LC}}$

Therefore

$f = \frac{1}{2\pi}\sqrt{\frac{1}{0.280 * 10^{-3}* 20 * 10^{-6}}}$

f = 2126.9 Hz

b)

U = 0.5*(20*10^{-6})(150)^2

U = 0.225 J

c)

150 = 0.28*10^{-3} dI/dt

dI/dt= 535714A/s

Read more about Work

https://brainly.com/question/756198