A circular core of high relative permeability material is shown in figure Q2c. Insulated wire is wrapped around the core to make an inductor of 100mH. An alternating current source is connected in series with a resistor to the magnetic circuit. The supplied current is sinusoidal with a RMS (Root Mean Square) current of 10 mA at a frequency of 12kHz. i) Determine the RMS voltage across the inductor. The AC current source is replaced by a DC current source. The output is initially isolated from the magnetic circuit by an open switch. The DC supply is set to 10 mA. The switch is then closed and the current is allowed to flow.

Answers

Answer 1

RMS voltage across the inductor is determined using the formula for RMS voltage across an inductor, Vrms = 2πfLI where L is the inductance of the inductor and I is the RMS current supplied to the inductor. The frequency f is given to be 12 kHz, and the inductance L is given to be 100 mH.Explanation :The inductor has an inductance L = 100 mHThe frequency f is given to be 12 kHz

The RMS current supplied to the inductor is I = 10 mA.The formula for RMS voltage across an inductor is given as:Vrms = 2πfLIOn substituting the values of L, f and I in the formula, we get,Vrms = 2π(12 kHz)(100 mH)(10 mA) = 7.54 Vii) The magnetic circuit has a circular core of high relative permeability material and an insulated wire wrapped around the core to form an inductor. A DC current source is connected to the circuit. Initially, the output of the DC current source is isolated from the magnetic circuit by an open switch.

Then, the DC supply is set to 10 mA and the switch is closed. As the switch is closed, the current starts to flow in the circuit through the inductor. However, since the core is made up of a high permeability material, the inductance of the circuit increases. This leads to a delay in the buildup of current in the circuit. Due to the delay, the current in the circuit does not reach the steady-state immediately after the switch is closed.

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Related Questions

Two point charges, Q1 and Q2, are located at (1, 2, 0) and (2, 0, 0), respectively. Find the relation between Q1 and Q2 such that the force on a test charge at the point P(-1, 1, 0) will have

a. No x-component

b. No y-component

Answers

the relation between Q1 and Q2 such that the force on the test charge at P will have no x-component is Q1 * sqrt(10) = Q2 * sqrt(13).

To find the relation between Q1 and Q2 such that the force on a test charge at the point P(-1, 1, 0) will have no x-component, we need to set up an equation using Coulomb's law. Coulomb's law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

Let's assume Q1 is the charge at (1, 2, 0) and Q2 is the charge at (2, 0, 0). The equation for the force on the test charge at P(-1, 1, 0) is:

F = k * (Q1 * Q2) / r^2

where k is the Coulomb's constant and r is the distance between the test charge and the point charge.

Now, since we want the force to have no x-component, we can set the x-components of the forces from Q1 and Q2 equal to each other. Let's denote the distance between Q1 and P as r1 and the distance between Q2 and P as r2.

The x-component of the force from Q1 is given by:

F1x = k * (Q1 * Q) / r1^2

The x-component of the force from Q2 is given by:

F2x = k * (Q2 * Q) / r2^2

Since we want F1x = F2x, we can equate the two equations:

k * (Q1 * Q) / r1^2 = k * (Q2 * Q) / r2^2

Canceling out the constants and rearranging the equation, we get:

(Q1 * Q) / r1^2 = (Q2 * Q) / r2^2

Now, substituting the values for r1, r2, and the coordinates of P, we have:

(Q1 * Q) / (sqrt(2^2 + 3^2)) = (Q2 * Q) / (sqrt(3^2 + 1^2))

Simplifying further:

(Q1 * Q) / sqrt(13) = (Q2 * Q) / sqrt(10)

Cross multiplying:

(Q1 * Q * sqrt(10)) = (Q2 * Q * sqrt(13))

Dividing both sides by Q * Q:

Q1 * sqrt(10) = Q2 * sqrt(13)

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If you charge your céll phone battery with 100 units of energy and you get 120 units of energ) out of the battery, which law of thermodynamics are you violating? 1. A) \( 0^{\text {th }} \) Law

Answers

The situation you described, where you charge your cellphone battery with 100 units of energy and get 120 units of energy out of the battery, violates the First Law of Thermodynamics, also known as the Law of Conservation of Energy.

The First Law of Thermodynamics states that energy cannot be created or destroyed within an isolated system. It can only be converted from one form to another or transferred between different parts of the system. In simpler terms, the total amount of energy in a closed system remains constant.

In the case of your cellphone battery, charging it with 100 units of energy and extracting 120 units of energy would imply that the battery is generating energy on its own, which contradicts the principle of conservation of energy. Such a violation would imply the creation of energy from nothing, which is not possible according to our current understanding of physics.

Therefore, the situation you described goes against the First Law of Thermodynamics. In reality, there are losses associated with energy conversions and transfers, such as heat dissipation and inefficiencies in the charging and discharging processes, which would prevent you from obtaining more energy from the battery than you put into it.

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An electron beam is diffracted from the crystal plane (220) of cubic sample with lattice constant 5.43 A at diffraction angle 3.66^ ∘. What is the energy of the incident electrons.

Answers

The energy of the incident electrons is 1.213 keV.

Diffraction angle = θ = 3.66°

Lattice constant = a = 5.43 A= 5.43 × 10⁻¹⁰ m

Crystal plane = (220)

For a cubic crystal, the atomic spacing between (hkl) planes is given as

a/hkl = [tex]\sqrt{2}[/tex] / hkl ---(1)

The energy of an electron beam is given as

E = (hc) / λ ---(2)

where

h is Planck's constant, c is the velocity of light, and λ is the wavelength of the electron beam.

The Bragg's law for diffraction is given as

kλ = 2d sinθ ---(3)

where k is a positive integer

Now we need to find the energy of the incident electrons.

From equation (1), the atomic spacing between (220) planes is

a/220 =  [tex]\sqrt{2}[/tex] / 220 = 5.43 × 10⁻¹⁰ / [tex]\sqrt{2}[/tex]

Using equation (3), the wavelength of the incident electron beam is given as

kλ = 2d sinθ = 2 x 5.43 × 10⁻¹⁰ /[tex]\sqrt{2}[/tex] × sin 3.66°

= 1.631 × 10⁻¹⁰ m

Now using equation (2), the energy of the incident electrons is

E = (hc) / λ = (6.626 × 10⁻³⁴ Js) × (3 × 10⁸ m/s) / 1.631 × 10⁻¹⁰ m

= 1213 eV

= 1.213 keV

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