When an electric field is applied to a conductor, there occurs a large scale physical movement of free electrons because these are available in large numbers in Conductor.

On the other hand, if an electric field is applied to an insulator, there is hardly any movement of free electrons because these are just not available in an insulator. Plastics, wood, and rubber are examples of good insulators. Pure water is also an insulator. Tap water, however, contains salts that form ions which can move through the liquid, making it a good conductor.

The insulator is also called the dielectric. There are practically no free electrons in the dielectric. The electrons in dielectric normally remain bounds to their respective molecules.

There are some materials, called semiconductors, which are intermediate between conductors and insulators.

Superconducting materials are the materials which conduct electricity without resistance below a certain temperature. Superconductivity is one of the most exciting phenomena in Physics, because of the peculiar nature and the wide application of this phenomenon. This phenomenon of superconductivity was first discovered by a Dutch physicist, H.K. Onnes. Superconducting materials are having very good electrical and magnetic properties. Before the discovery of superconductors, it is believed that the electrical resistivity of the material becomes zero, only at the absolute temperature.

Given

Resistance R = 5Ω

Current I = 2 A

Power dissipated by the resistor is

P = I²R

P = 2² × 5

P = 20 watts

An inductor is a device which temporarily stores energy in the form of the magnetic field. It is usually a coil of wire. One of the basic property of the electromagnetism is that when you have current flowing through the wire it creates a small magnetic field around it.

One current first start to flow through the inductor a magnetic field start to expand then after some time magnetic field becomes constant then we have some energy stored in the magnetic field.

Once a constant magnetic field is generated in the Inductor, it will not change any further. As magnetic flux = N x I (Turns x Current), Inductor will draw a constant current to maintain the magnetic field.

Once current stop flowing the magnetic field start to collapse and the magnetic energy turned back into electric energy.

So when the current flowing through an inductor changes, the magnetic field also changes in the inductor and emf (electromotive force) is induced in the inductor as per Faraday’s law of electromagnetic induction.

According to Lenz’s law, the direction of electromotive force(emf) opposes the change of current that created it.  V= -Lx dI/dt (rate of change of current)

So inductor opposes any change of current through them.

The resistance of the conductor is determined by the 

R = ρL/A

Where

ρ = Resistivity of the conductor

L = Length of the conductor = 10 m

A = Area of the conductor = πR² = 3.14 × 0.2² = 0.125 m²

R = Resistance = 2 Ω

Therefore the resistivity is

ρ = A × R ⁄ L

= 0.125 × 2 ⁄ 10

ρ = 0.025 Ω-m

The Farad is the practical and the Sl unit of capacitance. The unit, named after Michael Faraday (1791-1867), was first suggested by Latimer Clark in 1867. The capacitor has a capacitance of 1 farad when a charge of 1 coulomb raises the potential between its plates to 1 volt.

The S.I unit of Inductance is Henry.

The S.I unit of resistance is OHM.

The S.I unit of Reluctance is amp-turns/Weber or Henry⁻¹

 

In the given circuit the capacitance C₁ and C₂ are parallel with the capacitance C₃ i.e

(C₁ || C₂) + C₃

∴(20 × 30) ⁄ (20 + 30) + 20

C_A= 12 + 20 = 32 μF

Now capacitance C_A, C₄, & C₅ are in the series therefore

C_eqv = (1/30 + 1/20 + 1/20)

C_eqv = (60/8) = 7.5 μF

The power can be defined as

P = V² ⁄ R

Given

P = 200 W

V = 220 V

200 = 220² ⁄ R

R = 242 Ω

In the given circuit the resistance R₉ and R₈ are Parallel with Resistance R₁₀ therefore

(R₉ + R₈) || R₁₀

= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}

= (6 × 12) ⁄ (6 + 12)

R_A = 4Ω

Now resistance R_A and R₇ is parallel with Resistance R₆

∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

R_B = 4Ω

Now the circuit becomes as shown in the figure

Resistance R_B and R₇ is parallel with Resistance R₄

{(4 + 8) × 6} ⁄ {(4 + 8) + 6}

= (12 × 6) ⁄ (12 + 6)

R_C = 4Ω

Now resistance R_C and R₃ is parallel with Resistance R₂

{(4 + 2) × 6} ⁄ {(4 + 2) + 6}

R_D = 3

Now our final circuit becomes as shown in the figure

Therefore the equivalent resistance is

R₁ + R_D

Req = 4 + 3 = 7Ω

The resistance of the conductor is determined by the 

R = ρL/A

Let the resistance of the wire 1 be R₁
and the resistance of the wire 2 is R₂

The wire has the same length and resistivity therefore

The resistance of the conductor is determined by the 

ρ₁ = ρ₂
&
L₁ = L₂

Now the cross-sectional area of the first wire is 2 times the second wire

A₁ = 2A₂

Also, the resistance of the 2nd wire is 20 ohms.

$\begin{array}{l}\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{A_2}}}{{2{A_2}}}\\\\\dfrac{{{R_1}}}{{20}} = \dfrac{{{A_2}}}{{2{A_2}}}\\\\{R_1} = 10\Omega \end{array}$

The power can be defined as

P = I²R

Let the Power dissipated by Bulb A be

P = I²R_A = 100 = I²R_A

And power dissipated by Bulb B be

P = I²R_B = 10 = I²R_B

As we Know that  the current flow in the series circuit is same

R_A ⁄ 100 = R_B ⁄ 10

R_A ⁄ R_B = 10

or

R_A = 10R_B

Power can also be defined as the

P = V²/R

Total Power consumption = 100 Watt + 10 Watt = 110 Watt

Applied voltage = 20 Volt

∴ 110 = 20² ⁄ R

or

R = 20² ⁄ 110 = 40 ⁄ 11

Now in series connection, the equivalent resistance is the sum of the individual resistance

∴ R = R_A + R_B

40 ⁄ 11 = 10R_B + R_B

R_B = 0.33 Ω

Hence R_A = 10R_B

= 10 × 0.33 =3.3

R_A = 3.3Ω