(i) Kalaivani's mother took a small PVC pipe which has water in it and held in a horizontal position. Water came out drop by drop. When she raised one of the ends of the pipe, the water through the other end came out with a greater speed. She said initially when I was holding the pipe in a horizontal position, pressure at the both the ends are same. So water did not come out. The difference in pressure (Pressure difference) between the two ends is zero. But when I raised one of the ends, the top end has high pressure, and the bottom end has low pressure. Due to this pressure difference, water came out with a faster rate. So compare a conductor or a wire to this PVC pipe The wire is connected to a battery, whose one end has higher potential and the other end has lower potential. Due to this difference in potential, electrons from negative terminal (lower potential end) flow towards the positive terminal (higher potential end). Here electron flow is compared with the water flow and the electric potential is compared with the pressure. Kalaivani felt very happy, after hearing her mother's genius explanation.
(ii) When Kalaivani asked how to understand about the concept of resistance from her mother's scientific explanations, her mother continued. If the water pipe is smooth inside, then there is no charge in the speed of flow of water. i.e., zero resistance. But when the pipe has some blocks inside at different points, the water changes its speed of flow at different points, likewise, whenever electrons flow inside the conductor, they may collide with each other, due to which their speed will be resisted (blocked), which forms as electrical resistance. Her explanation made Kalaivani satisfied.
(iii) When two resistors of resistances, R₁ & R₂ are connected in series to a battery source.
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }\)
When R₁ & R₂ are connected in parallel,
\(\frac { 1 }{ { R }_{ p } } +\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \Rightarrow \frac { 1 }{ { R }_{ p } } =\frac { { R }_{ 2 }+{ R }_{ 1 } }{ { R }_{ 1 }{ R }_{ 2 } } \)
\({ R }_{ p }=\frac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } \)

(i) The earth wire provides a low resistance path to the electric current.
(ii) The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of metallic electric appliance.
(iii) Thus, the earth wire serves as a protective conductor, which saves us from electric shock.

Electric potential : The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.
Electric potential difference : The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.

R₁ = 1 Ω, R₂ = 2 Ω, R₃ = 4 Ω Current I₁ = 1 A
The potential difference across the 1 Ω resistor = I₁ R₁ = 1 × 1 = 1 V
Since, the resistors are connected in parallel in the circuit, the same potential difference will exist across the other resistors also.
So, the current in the 2 Ω resistor, \(\frac { V }{ { R }_{ 2 } } =\frac { 1 }{ 2 } =0.5A\)
Similarly, the current in the 4 Ω resistor,
\(\frac { V }{ { R }_{ 3 } } =\frac { 1 }{ 4 } =0.25A\)

Copper wire is used as connecting wires because copper has very low resistivity.

(i) Electric heater.
(ii) Electric iron work on the heating effect of current.

(i) Melting point of tungsten is very.
(ii) Hence it is used in a filament bulb.
(iii) But a fuse wire should be made up of material which has low melting point.

Given:
I = 5 A 
t = 1 s
Number of electrons?
\(\mathbf{q}=\mathbf{I\times t}=5 \times 1=5 \mathrm{C} \)
q = ne where n is the number of electrons; e is the charge of an electron which is equal to \(1.6 \times 10^{-19} \mathrm{C}\)
\(n=\frac{q}{e}=\frac{5}{1.6 \times 10^{-19}}=3.125 \times 10^{19}=31.25 \times 10^{18}\) electrons.

Given :
Charge passing through bulb, q = 30C
Current drawn by 100 W bulb, I = 680 x 10^-3 A
To find: Time taken to pass
through the bulb t =? t=\(\frac{q}{I}\)
Solution :
\(I=\frac { q }{ t } =\frac { 30 }{ 680\times { 10 }^{ -3 } } \)
=441.173
Time required, t =7.35 min

The fuse will melt and the circuit breaks if electric current more than the rating of fuse (i.e., 5 A) flows in the circuit. Electric current flowing in the circuit,
\(I=\frac { P }{ V } =\frac { 1.5\times 1000W }{ 220V } =6.82A\)
Since, current flowing in the circuit (6.82 A) is more than the rating of fuse (5 A), therefore, the fuse will melt and the electrical heater does not work. To operate the heater, fuse of rating 10 A is to be put in the circuit.