Parallel connection is used in domestic appliances.
Reason:
(i) Each appliance will get the full voltage.
(ii)The parallel circuit divides the current through the appliances.
(iii) Each appliance will get the proper current depending on its resistance.
(iv) Each of them can be put on / off independently.

According to Ohm's law, at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V' between the two ends of the conductor.
\(I\alpha V \Rightarrow\)  V = IR

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.

a) Wire is cut into 5 equal parts. Since all dimensions are same, resistance of each wire is equal and has a value = \(\frac {R}{5}\)
b) Formula for finding the effective resistance when connected in parallel is
\(\frac{1}{R_{p}^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}\)
Here, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}_{3}=\mathrm{R}_{4}=\mathrm{R}_{5}=\frac{\mathrm{R}}{5}\)
\(\frac{1}{R_{p}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}\)
\(\frac{1}{R_{p}}=\frac{25}{R}  \)
\(R_{p}=\frac{R}{25} \Omega\)
c) If the resistors are connected in series, then the effective resistance will be
\(\mathrm{R}_{\mathrm{s}}=\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5}+\frac{\mathrm{R}}{5} \)
\(\mathrm{R}_{\mathrm{s}}=\frac{5 \mathrm{R}}{5}=\mathrm{R}\)
Ratio of effective resistance in series connection to that of the parallel connection is
\(\frac{R_{s}}{R_{p}}=\frac{R}{R / 25}=\frac{25}{1}\Rightarrow R_s:R_p=25:1\)

Given:
Power of the first electric bulb \(=100 \mathrm{~W}=100 / 1000=0.1 \mathrm{~kW}\)
Time = 5 hours
Power of the second electric bulb \(=60 \ \mathrm{watt}=\frac{60}{1000}=0.06 \mathrm{~kW}\)
Total number of bulbs = 4,
∴ 4 x 0.06 = 0.24 kW
Time = 5 hours.
Energy consumed in the month of January = ?
Energy = Power x time
Energy consumed by the first bulb in a day = 0.1 x 5 = 0.5 kWh
Energy consumed by the four 60 W bulb in a day =0.06 x 4 x 5 = 1.2 kWh
Total energy consumed by both the bulbs = 0.5 + 1.2 = 1.7 kWh
Total energy consumed in the month of January = 31 x 1.7 = 52.7 kWh