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Dmitry_Shevchenko [17]

3 years ago

9

A 500 pF capacitor is charged up so that it has 10μC of charge on its plates. The capacitor is then quicklyconnected to a 10 H i

nductor. Calculate themaximum energystored in the magnetic field of the inductoras the circuit oscillates. What is the current through the inductor when the maximum energy stored in theinductor is reached?

1 answer:

klio [65]3 years ago

8 0

Answer:

Explanation:

Given that,

A capacitor of capacitance

C = 500pF

Charge on capacitor is

Q = 10μC

Capacitor is then connected to an inductor of inductance 10H

L = 10H

Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor

So energy stored in capacitor can be determined by using

U = ½CV²

Then, Q = CV

Therefore V = Q/C

U = ½ C • (Q/C)² = ½ C × Q²/C²

U = ½Q² / C

Then,

U = ½ × (10 × 10^-6)² / (500 × 10^-9)

U = 1 × 10^-4 J

U = 0.1 mJ

So the energy stored in this capacitor is transfers to the inductor.

So, energy stored in the inductor is 0.1mJ

B. Current through the inductor

Energy in the inductor is given as

U = ½Li²

1 × 10^-4 = ½ × 10 × i²

1 × 10^-4 = 5× i²

i² = 1 × 10^-4 / 5

i² = 2 × 10^-5

I = √(2×10^-5)

I = 4.47 × 10^-3 Amps

Then,

I = 4.47 mA

You might be interested in

Two obiect accumulated a charge of

tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

$q_1=4.5\mu C=4.5\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2.8\mu C=2.8\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

$F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N$

So, the closest answer is

A) 181.24 N

3 0

3 years ago

A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w

Sliva [168]

Answer:

$a = 0.894\ m/s^2$

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

$v_f=v_o+at$

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{8.94-4.47}{5}$

$a = 0.894\ m/s^2$

3 0

2 years ago

The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad

kenny6666 [7]

Answer:

a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,

c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

k = 2π / λ

λ = 2π / k

λ = 2π / 1.99 10⁷

λ = 3,157 10⁻⁷ m

w = 2π f

f = w / 2 π

f = 5.97 10¹⁵ / 2π

f = 9.50 10¹⁴ Hz

the wave speed is

v = 3,157 10⁻⁷ 9.50 10¹⁴

v = 2,9992 10⁸ m / s

b) The electric field is

Eo = 375 V / m

to find the magnetic field we use

E / B = c

B = E / c

B = 375 / 2,9992 10⁸

B = 1.25 10⁻⁶ T

c) The period is

T = 1 / f

T = 1 / 9.50 10¹⁴

T = 1.05 10⁻¹⁵ s

the wavelength value is

λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0

4 years ago

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

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

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

$z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}$

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

4 0

3 years ago

A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i

nalin [4]

<span>1.7 rad/s The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>

3 0

4 years ago