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2 years ago

A long time after the switch is closed, what is the voltage across the capacitor in terms of the supply voltage?

1 answer:

4 0

A long time after the switch is closed, the voltage across the capacitor in terms of the supply voltage is zero.

<h3>What is a Capacitor?</h3>

In an electric field, a capacitor is a device that stores electrical energy. It has two terminals and is a passive electrical component.

Capacitance refers to a capacitor's effect. While there is some capacitance between any two nearby electrical wires in a circuit, a capacitor is a component made to increase capacitance. The term "condenser" or "condensator" originally applied to the capacitor. Condenser microphones, sometimes known as capacitor microphones, are a remarkable exception to the general lack of usage of this name and its cognates in English.

Practical capacitors come in a wide variety of physical shapes and constructions, and there are numerous varieties that are used often. The majority of capacitors have two or more electrical conductors, frequently in the form of metallic plates or surfaces, spaced between by an insulating material. A conductor can be an electrolyte, thin film, metal bead that has been sintered, or foil. The capacitor's charge capacity is increased by the nonconducting dielectric. Glass, ceramic, plastic film, paper, mica, air, and oxide layers are a few examples of materials that are frequently employed as dielectrics. Many typical electrical gadgets use capacitors in their electrical circuits.

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A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3

4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

$B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT$

8 0

3 years ago

In lightning , light is seen first and sound is heard later it is due to

Maksim231197 [3]

Answer:

option 4

Explanation:

Light's velocity in air ( 3 × 10^8 m/s ) is much greater than sound's velocity in air ( 343 m/s )

Hence due to difference in velocities , during lightning light is seen first & sound is heard later

8 0

3 years ago

Read 2 more answers

Given:A=6x-2y B:-4x-8y C:-3x+9y. Commute A+B-C

DedPeter [7]

<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y

(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)

5x - 19y

8 0

3 years ago

Explain any two factors that affect the pressure at a point inside a liquid

vredina [299]

The density of the body and the height or the depth of the body since the formula of liquid pressure is density x height gravity

4 0

2 years ago

Read 2 more answers

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago