As the wire is bent in the form of a closed circle, it will be have like two resistors of 5\(\Omega \) each, connected in parallel. Therefore, the equivalent resistance can be calculated as:
\(\frac { 1 }{ R_{ eq } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ R_{ eq } } =\frac { 1 }{ 5 } +\frac { 1 }{ 5 } \)=\(\frac { 1 }{ R_{ eq } } =\frac { 2 }{ 5 } \)
\(\frac { 1 }{ R_{ eq } } =\frac { 5 }{ 2 } =2.5\Omega \)
The effective resistance between the two points at the ends of an diameter of the circle is 2.5 \(\Omega \).

Nichrome should be used as a heating element in an electric iron because its melting point is higher than iron and silver.

(i) Electric power is related to voltage and electric current as:
Power= Voltage x Current
It is given that the rating of the electric iron is 750W, 220V.
Let the current passing through the bulb be I.
Therefore, 
750=220 x I
I=\(\frac{750}{220}\)=3.41A
Therefore, 3.41A current passes through the electric iron.
It is given that the rating of the electric bulb is 750W,220 V
Therefore, P=\(\frac { { V }^{ 2 } }{ P } \)=\(\frac { { 220 }^{ 2 } }{ 750 } \)=64.53\(\Omega \) when in use.
(ii) Electric power is related to voltage and resistance as:
\(\frac { { \left( Voltage \right) }^{ 2 } }{ Resistance } \)
That is,
P=\(\frac { { V }^{ 2 } }{ R } \)
=64.53\(\Omega \)

Voltage (V) = 5.0 V
Current (I) = 100 mA = 0.1 A
(i) Power = VI = 5\(\times\)0.1= 0.5W
(ii) Resistance = \(\frac { V }{ I } =\frac { 5 }{ 0.1 } =50\Omega \)

Coulomb

The resistance of 25 W lamp bulb has a higher value as compound to that of 50 W lamp bulb. Resistance of the 25 W lamp bulb. Resistance of the 25 W lamp bulb will be double of that of the 50 W lamp bulb.

Watt

The potential difference between two points is said to be 1 V if 1 J of work is done in moving 1 Coulomb of electric charge from one point to other point.

Electric cell or battery is a device that helps to maintain a potential difference across a conductor.

We know that, charge on one electron = 1.6 \(\times\) 10^-19C
\(\Rightarrow\) 1.6 \(\times\) 10^-19 coulomb charge = 1 electron.
\(\therefore\) 1 coulomb charge=\(\frac{1}{1.6\times 10^{-19}}\simeq 6.25\times 10^{18}\) electrons

The SI unit of electric current is ampere (A).
The current flowing through a conductor is said to be 1A, if a charge of 1coulomb (C) flows through it in 1second (s)
or \(1 A=\frac{1 C}{1 s}\)

The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. Electric current flows in opposite direction to the movement of electrons.

(ii) When the key is closed, total voltage applied across the circuit will be = 2 ✕ 5V = 10 V
     Total resistance (R) applied across the circuit in Ohms = 5+10+15= \(30\Omega \) 
     By Ohm's Law,
     Current (I) = \(\frac { V }{ R } \)
       \(I=\frac { 10 }{ 30 } =0.33A\)

(a) Ohm's Law states that at constant temperature, current flowing through a conductor is directly proportional to the       potential difference across its ends.

A number of resistors are said to be connected in parallel if one end of each is connected to one point and the other end is connected to another point so that the potantial difference across each resistor is the same and is equal to the applied potential difference between the two points. If we connect a voltmeter first across R₁, then across R₂ ad lastly across R₃, it will show same voltage.