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3 years ago

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A capacitor is connected to an ac generator that has a frequency of 3.0 kHz and produces a rms voltage of 2.0 V. The rms current

in the capacitor is 31 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.8 kHz, the rms current in the capacitor is 74 mA. What rms voltage does the second generator produce

1 answer:

Dimas [21]3 years ago

7 0

Answer:

In second case voltage will be 2.986 volt

Explanation:

In first case we have given frequency f = 3 kHz = 3000 Hz

Voltage v = 2 volt

Current i = 1 mA

So capaitive reactance $X_C=\frac{V}{i}=\frac{2}{31\times 10^{-3}}=64.5161ohm$

$X_C=\frac{1}{\omega C}$

So $64.5161=\frac{1}{2\times 3.14\times 3000\times C}$

$C=8.22\times 10^{-7}F$

In second case frequency f =4.8 kHz = 4800 Hz

Capacitive reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 4800\times 8.22\times 10^{-7}}=40.3577ohm$

Current i = 74 mA

So voltage $V=iX_C=74\times 10^{-3}\times 40.3577=2.986volt$

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Complete Question

An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick

Answer:

The largest total area of the oil slick $A = 8.257 *10^{9} \ m^2$

Explanation:

From the question we are told that

The volume of oil the escaped is $V = 18000 \ L$

The refractive index of oil is $n_o = 1.1$

The refractive index of water is $n_w = 1.33$

The wavelength of the light is $\lambda = 485 \ nm = 485 * 10^{-9} \ m$

Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as

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It is usually take as 1 unless stated otherwise by the question

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$d = 1 * \frac{485 *10^{-9}}{2 * 1.1}$

$d = 218 nm$

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

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hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

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