(a) (i) Define Electric Power and write its SI unit. (ii) Two bulbs rated 100 W; 220 V and 60 W; 220 V are connected in parallel to an electric mains of 220 V. Find the current drawn by the bulbs from the mains. OR (b) (i) State Joule's law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t. (ii) A 5 $\Omega$ resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 s.

Option (a)

(i) Electric Power:

Electric power is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit.

Mathematically,

$$P = V I = I^2 R = \frac{V^2}{R}$$

where $V$ is the potential difference, $I$ is the current, and $R$ is the resistance.

• SI Unit: The SI unit of electric power is watt (W).

(ii) Numerical Solution:

Since the two bulbs are connected in parallel to the $220\ \text{V}$ mains, the potential difference across each bulb is $220\ \text{V}$.

• Current drawn by the $100\ \text{W}$ bulb ($I_1$):

$$I_1 = \frac{P_1}{V} = \frac{100\ \text{W}}{220\ \text{V}} = \frac{5}{11}\ \text{A} \approx 0.45\ \text{A}$$

• Current drawn by the $60\ \text{W}$ bulb ($I_2$):

$$I_2 = \frac{P_2}{V} = \frac{60\ \text{W}}{220\ \text{V}} = \frac{3}{11}\ \text{A} \approx 0.27\ \text{A}$$

• Total current drawn from the mains ($I$):

$$I = I_1 + I_2 = \frac{5}{11}\ \text{A} + \frac{3}{11}\ \text{A} = \frac{8}{11}\ \text{A} \approx 0.73\ \text{A}$$

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OR

Option (b)

(i) Joule's Law of Heating:

Joule's law of heating states that the heat ($H$) produced in a resistor is:

1. Directly proportional to the square of current ($I^2$) for a given resistance,
2. Directly proportional to the resistance ($R$) for a given current, and
3. Directly proportional to the time ($t$) for which the current flows through the resistor.

• Mathematical Expression:

$$H = I^2 R t$$

Using Ohm's law ($I = V/R$), this can also be expressed as:

$$H = \frac{V^2}{R} t$$

(ii) Numerical Solution:

Given:

• Resistance, $R = 5\ \Omega$
• Potential difference, $V = 6\ \text{V}$
• Time, $t = 10\ \text{s}$

Using the formula for heat energy:

$$H = \frac{V^2}{R} t$$

$$H = \frac{6^2}{5} \times 10 = \frac{36}{5} \times 10 = 36 \times 2 = 72\ \text{J}$$

Answer: The energy dissipated as heat in $10\ \text{s}$ is $72\ \text{J}$.