Given, Resistance, R = 8 \(\Omega\) , Current, J = 15 A
Time, t = 2 h = 7200 s
\(\therefore\) Heat developed. H = J²Rr
=15 \(\times\)15 \(\times\) 8 \(\times\) 7200 J
\(\therefore\) Rate of heat developed.
\(P=\frac{H}{t}=\frac{15\times 15\times 8 \times 7200}{7200}\)
= 1800 W or 1800J/s
Thus, the rate at which heat is developed in the heater is 1800 joule per second.

Given, P₁ = 250 W, P₂ = 1200 W,
t₁ = 1h = 3600 s, t₂ = 10 min = 600 s
\(\therefore\) Energy
Q₁ = P₁t₁ = 250 \(\times\)3600 = 900000 J = 900 kJ
and Q₂ = P₂t₂ = 1200 \(\times\) 600 = 720000 J = 720 kJ
Thus, TV set uses more energy.

A voltmeter is always connected in parallel in the circuit to measure the potential difference between two points.

 A fuse used must be of current capacity less than the maximum current which a circuit or on appliance can with stand. In other words, if a current larger than a specified value flows in a circuit, temperature of fuse wire increases to its melting point. The fuse wire melts and the circuit breaks.

The number of electrons flowing through the circuit will remain constant. Therefore the current flowing will also be constant. The number of electrons have to travel in a fixed path. Therefore, the value of current is same at each and every point.

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

(i) Resistance of a glowing lamp is related to its power and voltage as
Power=\(\frac { { \left( voltage \right) }^{ 2 } }{ Resistance } \)
or, P=\(\frac { { V }^{ 2 } }{ R } \)
Therefore, R=\(\frac { { V }^{ 2 } }{ P } \)=\(\frac { { 220 }^{ 2 } }{ 100 } \)=484\(\Omega \)
Therefore, the resistance of the bulb when glowing is 484\(\Omega \)
(ii) Power=100 W=0.1 kW
Energy= Power x time
=0.1 kW x 5h
=0.5 kWh
0.5 kWh is the amount of energy is consumed by the bulb per day.

(i) Power = Voltage Current
              = 220\(\times\)3.4
              = 748 Watt
(ii) Power = Voltage / Resistance
           P = V² / R
           R = V² / P=220² / 748 = 64.71 \(\Omega \)

Given, I = 5 A, V = 220 V, t = 2h
\(\therefore\) Power of motor,
P = VI = 220 \(\times\)5 = 1100 W = 1.1 kW
\(\therefore\) Energy consumed = P t = 1.1 \(\times\)2 = 2.2 kWh
Thus, the power of the motor is 1.1 kW and energy consumed is 2.2 kWh.

Let resistance of lamp, R₁ = 100 \(\Omega\)
Resistance of toaster, R₂ = 50 \(\Omega\)
Resistance of filter, R₃ = 500 \(\Omega\)
Net resistance,
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
[\(\because\) R₂, R₂ and R₃ are connected in parallel]
\(\frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{16}{500}\) or \(R=\frac{500}{16}=31.25\Omega \)
So, resistance of iron to take same current as much Current drawn by all the appliances should be 31.25 \(\Omega\) . Current through circuit,
\(I=\frac{V}{R}=\frac{220}{31.25}=7.04 A\)
Thus, current through iron is 7.04 A.

Resistance is inversely proportional to the area of cross-section of the wire. Since, thìck wire has a large area of cross-section, its resistance will be less. Thus, current will flow more easily through the thick wire.

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 \(1 A=\frac{1 C}{1 s}\)