Question

A bulb is rated at 200 V, and 100 W. What is its resistance? Five such bulbs are lit for 4 hours. How much electrical energy is consumed? Calculate the cost if the rate is 50 paisa per unit.

Answer 1

Concept used: Electric current

Prerequisite: Formula electrical power

Step 1: Given data,

1. The voltage rate of bulb is V=200 V.
2. The power rate of bulb is P=100 W.
3. The time taken is t=4 hours.
4. The number of bulbs is n=5.
5. The unit price of electricity is c=50 paisa per unit.

Step 2: Calculate the resistance of the bulb,

The formula of power in terms of resistance and voltage is given by,

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

Here, P is the voltage, V is the voltage, and R is the resistance.

Substitute the given values in above equation,

$\begin{array}{rcl}100\mathrm{W}& =& \frac{{(200\mathrm{V})}^2}{R}\\ R& =& 400\mathrm{\Omega }\end{array}$

Thus, the resistance of the bulb is 400 Ω.

Step 3: Calculate the electric energy used,

The current passed through the bulb can be calculated by Ohm's law as,

$$\begin{gathered} I=\frac{V}{R} \\ I=\frac{200\mathrm{V}}{400\mathrm{\Omega }} \\ I=0.5\mathrm{A} \end{gathered}$$

The expression for the electricity consumed by the bulb is given by,

$E=n{I}^{2}Rt$

Substitute the given values in the above equation,

$\begin{array}{rcl}E& =& 5{\left(0.5\mathrm{A}\right)}^2\left(400\mathrm{\Omega }\right)\left(4\mathrm{hrs}\right)\\ E& =& 2000\mathrm{Wh}\times \frac{1\mathrm{kWh}}{1000\mathrm{Wh}}\\ E& =& 2\mathrm{kWh}\end{array}$

Thus, the electrical energy consumed is 2 kWh.

Step 4: Calculate the cost,

The cost of the electric energy is given by,

$$\begin{gathered} C=cE \\ C=\left(50\frac{\mathrm{paisa}}{\mathrm{unit}}\right)\left(2\mathrm{kWh}\times \frac{1\mathrm{unit}}{1\mathrm{kWh}}\right) \\ C=100\mathrm{paisa}\times \frac{1₹}{100\mathrm{paisa}} \\ C=₹1 \end{gathered}$$

Thus, the cost of electrical energy is ₹ 1.