(A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R). (ii) An electric oven is designed to work on the mains voltage of 220 V. This oven consumes 11 units of electrical energy in 5 hours. Calculate: (a) power rating of the oven (b) current drawn by the oven (c) resistance of the oven when it is red hot OR (B) (i) Write the relation between resistance R and electrical resistivity $\rho$ of the material of a conductor in the shape of cylinder of length $l$ and area of cross-section A. Hence derive the SI unit of electrical resistivity. (ii) The resistance of a metal wire of length $3\mathrm{m}$ is $60\Omega$ . If the area of cross-section of the wire is $4\times 10^{-7}\mathrm{m}^2$ , calculate the electrical resistivity of the wire. (iii) State how would electrical resistivity be affected if the wire (of part 'ii') is stretched so that its length is doubled. Justify your answer.

OPTION (A)

(i)

• Electric Power: It is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit.
• Expression in terms of $V$ and $R$:

We know that electric power is given by:

$$P = V I$$

From Ohm's law, $I = \frac{V}{R}$. Substituting this value in the power equation:

$$P = V \left(\frac{V}{R}\right) = \frac{V^2}{R}$$

(ii)

Given:

• Voltage, $V = 220\text{ V}$
• Electrical energy consumed, $E = 11\text{ units} = 11\text{ kWh}$
• Time, $t = 5\text{ hours}$

• (a) Power rating of the oven ($P$):

$$P = \frac{E}{t} = \frac{11\text{ kWh}}{5\text{ h}} = 2.2\text{ kW} = 2200\text{ W}$$

• (b) Current drawn by the oven ($I$):

$$P = V I \implies I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}$$

• (c) Resistance of the oven when it is red hot ($R$):

$$R = \frac{V}{I} = \frac{220\text{ V}}{10\text{ A}} = 22\ \Omega$$

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OPTION (B)

(i)

• Relation: The resistance $R$ of a uniform metallic conductor is directly proportional to its length ($l$) and inversely proportional to its area of cross-section ($A$):

$$R = \rho \frac{l}{A}$$

where $\rho$ is the electrical resistivity of the material.

• Derivation of SI Unit of Resistivity:

Rearranging the formula for resistivity:

$$\rho = \frac{R \cdot A}{l}$$

Substituting the SI units of the respective physical quantities:

$$\text{SI unit of } \rho = \frac{\text{ohm } (\Omega) \times \text{meter}^2\ (\mathrm{m}^2)}{\text{meter } (\mathrm{m})} = \Omega\ \mathrm{m}$$

Thus, the SI unit of electrical resistivity is ohm-meter ($\Omega\ \mathrm{m}$).

(ii)

Given:

• Length of the wire, $l = 3\text{ m}$
• Resistance, $R = 60\ \Omega$
• Area of cross-section, $A = 4 \times 10^{-7}\text{ m}^2$

Using the formula:

$$\rho = \frac{R \cdot A}{l}$$

$$\rho = \frac{60\ \Omega \times 4 \times 10^{-7}\text{ m}^2}{3\text{ m}}$$

$$\rho = 20 \times 4 \times 10^{-7}\ \Omega\ \mathrm{m} = 8 \times 10^{-6}\ \Omega\ \mathrm{m}$$

(iii)

• Effect on Electrical Resistivity: The electrical resistivity of the wire will remain unchanged (i.e., it will remain $8 \times 10^{-6}\ \Omega\ \mathrm{m}$).
• Justification: Electrical resistivity is a characteristic property of the material of the conductor. It depends only on the nature of the material and its temperature, and is completely independent of the geometrical dimensions (length and area of cross-section) of the conductor.