Question

In a house, electricity is supplied at 200 volts. Two electric bulbs of resistance 100 \(\Omega\) and 200 \(\Omega\) are used daily in the house for 1 hour. Daily consumption of electric power is:

• A. 500 Wh
• B. 600 Wh
• C. 300 Wh
• D. 900 Wh

Hint:

To calculate energy consumption, use the formula \( E = P \times t \), where \(P\) is power in watts, and \(t\) is time in hours.

Answer 1

The Correct Option is C

The power consumed by an electric bulb is given by the formula: \[ P = \frac{V^2}{R} \] where \(P\) is the power, \(V\) is the voltage, and \(R\) is the resistance of the bulb.
Step 1: Power consumed by the first bulb (100 \(\Omega\)):
For the first bulb, resistance \(R_1 = 100 \, \Omega\), and the voltage supplied is \(V = 200 \, \text{V}\). \[ P_1 = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \, \text{W} \] This is the power consumed by the first bulb in 1 hour.
Step 2: Power consumed by the second bulb (200 \(\Omega\)):
For the second bulb, resistance \(R_2 = 200 \, \Omega\), and the voltage supplied is \(V = 200 \, \text{V}\). \[ P_2 = \frac{(200)^2}{200} = \frac{40000}{200} = 200 \, \text{W} \] This is the power consumed by the second bulb in 1 hour.
Step 3: Total power consumption.
The total power consumption is the sum of the power consumed by both bulbs: \[ P_{\text{total}} = P_1 + P_2 = 400 \, \text{W} + 200 \, \text{W} = 600 \, \text{W} \] This is the total power consumed in 1 hour. The total energy consumption in watt-hours (Wh) is: \[ E = P_{\text{total}} \times t = 600 \, \text{Wh} \]
Step 4: Conclusion.
The total power consumed daily by both bulbs is 600 Wh, which corresponds to option (B).