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gizmo_the_mogwai [7]
2 years ago
11

Find the rms current delivered by the power supply when the frequency is very large. Answer in units of A.

Physics
1 answer:
Verizon [17]2 years ago
8 0

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

I=\dfrac{V}{R}

Where, V = voltage

R = resistance

Put the value into the formula

I=\dfrac{47}{151}

I= 0.3112 \ A

Hence, The rms current is 0.3112 A.

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Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

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Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

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Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
3 years ago
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