Power P = 100 W and Voltage V = 200 V
Power P = V I
So, Current, \(I=\frac { P }{ V } =\frac { 100 }{ 200 } =0.5A\)
Resistance, \(R=\frac { V }{ I } =\frac { 200 }{ 0.5 } =400\Omega \)

i) The electricity produced in power stations is distributed to all the domestic and industrial consumers through overhead and underground cables.
ii) In our homes, electricity is distributed through the domestic electric circuits wired by the electricians.
iii) The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.
Main Box Contains:
a) Fuse Box:
i) The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
ii) The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current.
iii) An MCB is a switching device, which can be activated automatically as well as manually. It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit.
iv) It has a spring attached to the switch, which is attracted by an electromagnet when an excess current passes through the circuit.
v) Hence, the circuit is broken and the protection of the appliance is ensured.
b) Meter:
i) The meter is used to record the consumption of electrical energy.
Insulated Wire:
i) The electricity is brought to houses by two insulated wires.
ii) Out of these two wires, one wire has a red insulation and is called the "live wire'.
iii) The other wire has a black insulation and is called the 'neutral wire'.
iv) Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
v) After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required.
vi) After the main switch, these wires are connected to live wires of two separate circuits.
5A rating circuit:
i) Out of these two circuits, one circuit is of a 5 A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans.
15 A rating circuit:
i) The other circuit is of a 15 A rating, which is used to run electric appliances with a high power rating, such as air-conditioners, refrigerators, electric iron and heaters.
ii) It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
iii) One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.
iv) The electricity supplied to your house is actually an alternating current having an electric potential of 220 V.

a) Electric current:
(i) Electric current is often termed as 'current' and it is represented by the symbol 'I'.
(ii) It is defined as the rate of flow of charges in a conductor.
(iii) This means that the electric current represents the amount of charges flowing in any cross-section of a conductor (say a metal wire) in unit time.
(iv) If a net charge 'Q' passes through any cross-section of a conductor in time 't', then the current flowing through the conductor is
\(I=\frac{Q}{t}\)
b) SI unit of electric current:
(i) The SI unit of electric current is ampere (A).
(iii) The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,
\(1 \text { ampere }=\frac{1 \text { coulomb }}{1 \text { second }}\)
c) Ammeter is used to measure the electric current. It should be connected in series in a circuit.

Given:
\(\mathrm{R}=10 \Omega\) ; original length = l, new length l' = 3l; Area will decrease by 3 times.
\(R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{31}{A / 3}=\frac{9 \rho l}{A}=9 R=9(10)=90 \Omega\)

Given:
\(\mathrm{R}_{\mathrm{P}}=2 \Omega \)
\(R_{S}=9 \Omega \)
\(\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \)
\(R_{s}=R_{1}+R_{2} \)
\(\mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{1} +\mathrm{R}_{2}}{\mathrm{R}_{1}\mathrm{R}_{2}}\Rightarrow \frac{1}{2}=\frac{\mathrm{R}_{1} +\mathrm{R}_{2}}{\mathrm{R}_{1}\mathrm{R}_{2}} \Rightarrow R_1R_2=2(R_1-R_2) ......(1)\)
\(R_s=R_1+R_2=9 \Omega \Rightarrow R_1= 9-R_2 .....(2)\)
Substitute equation (2) in equation (1)
(9 - R₂) R₂  = 2(9 - R₂ + R₂)
\(9R-R_2^2 = -9r+18=0\frac{18}{-6-3}\)
(R₂ - 3)(R₂ - 6) = 0
R₂ = 3.6
from equation (1) and (2)
\((i) If \ \mathbf{R}_{2}=3 \Omega, \ \mathrm{R}_{1}=6 \Omega \)
\((ii) If \ R_{2}=6 \Omega, \ R_{1}=3 \Omega\)