The difference between the initial electric potential and the final potential measurements is known as potential difference and is measured in joules per coulomb, or volts. Thus, the potential difference is commonly referred to as the voltage.

or

Electric Potential is the measure of electrostatic potential energy per unit charge for a given point in space.

or

The difference in electric charge creates a potential difference or voltage. The size of the voltage is directly proportional to the charge difference.

Factors Affecting the Resistance

The resistance R offered by a conductor depends on the following factors :

1. Length of the material (l): The resistance of a material is directly proportional to the length. The resistance of the longer wire is more.
2. Cross-Section Area (a): The resistance of a material is inversely proportional to the cross-sectional area of the material. The more cross-sectional area allowed the passage of more number of electrons offering less resistance.
3. Nature of Material: As discussed earlier the conductor has a large number of free electrons hence it offers less resistance whereas Inductor has less number of free electrons hence which offers more resistance.
4. Temperature: The temperature of the material affects the value of the resistance. In the General case, the resistance of the material increases as its temperature increases.

In a series circuit, there is only one path in which current can flow. There are no branches through which current can leave, or enter the series loop. Since no additional current can enter or leave the circuit. the current is the same everywhere in a series circuit. This means the current is the same through every load in series

Mathematically 

I_T = I₁ = I₂ = I₃ ——— = I_N

Conductance

The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. The inverse quantity is electrical conductance and is the ease with which an electric current passes. Electrical conductance is measured in siemens (S) σ and denoted by (G).

Given

Current I = 2A

Voltage V = 40 V

Resistance R = V/I = 40/2 = 20

Hence Conductance

G = 1/R = 1/20 

G = 0.05 (Mho)

The three capacitor C₁, C₂, C₃ is connected in a series as shown in the figure

Now the Equivalent capacitance connected in the series will be

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

1/C_eq = 1/0.04 + 1/0.08 + 1/0.02

1/C_eq = 25 + 12.5 + 50

1/C_eq = 87.5

C_eq = 1/87.5

C_eq = 0.011

Given

Length =14 m

Radius = Diameter/2 = 0.6/2 =0.3m

Area of circle = πR² = π × 0.3²

Conductivity = 1/Resistivity

Resistivity = 1/12

The formula of resistivity is given as

∴ Resistance = (L × Ρ) ⁄ A

$.\begin{array}{l}R = \dfrac{1}{{12}} \times \dfrac{{14}}{{\pi \times {{0.3}^2}}}\\\\R = 14.17\Omega \end{array}$

Resistor values can be known by the color band.

0 black

1 brown

2 red

3 orange

4 yellow

5 green

6 blue

7 violet

8 grey

9 white

First color = digit
Second color=digit
Third color= power of ten
Fourth color =tolerance

Now in the given question 50 × 10⁰  ± 2%

5 = Green

0 = Black

10⁰ = Black

± 2% = Red

Resistance R = p(L/A)………..(1)

p is the resistivity of the material of wire.

L is the length of wire.

A is the area of cross-section of wire.

We multiply and divide equation (1) by l, then

R=p(L²/LA)

We further know that,

Volume = Length × Cross-Sectional Area

We can assume that the material volume does not change (density of the material not affected)

∴ R ∝ L².

Then, if we double the length, Resistance becomes R’=4R.

Interpretation can be drawn qualitatively as follows:

Due to an increase in the length of the wire, it will become thinner and longer. If it is thinner it will be harder for the charge to move through, so the resistance will increase. If it is longer, the charge has further to travel, so its resistance has increased.

Now in the given question, the 20Ω is resistance is stretched to double hence the new resistance will be

R’ = 4R = 4 x 20 = 80Ω

Now the resistance 80Ω is cut into two equal parts and are connected in the parallel circuit as shown in the figure

R = 40 || 40 = (40 × 40) ⁄ (40 + 40)

R = 20 Ω

Hence current I = V/R = 4/20

I = 0.2 A

If current flows for time t then the production of heat is given as

H = I²R.t

Where

I = current = 1 A

R = Resistance = 10 ohms

t = time = 5 sec

H = 1² × 10 × 5

H = 50 watt-sec or Joule