Correct Answer - Option 1 : 960 Ω
Concept:
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Electric bulb: It is an electric device that converts electric energy into heat and light energy.
- Resistance of the bulb \(R = \frac{{{{\left( {Rated\;voltage} \right)}^2}}}{{\left( {Rated\;power} \right)}}\)
- If we consider the bulb as a resistor then, we can easily find the current and voltage drops.
- \(Power\;consumed = {i^2}R\)
- Power consumed ∝ Brightness
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Loads connected in parallel: The voltage across each load connected in parallel is the same.
- The total current is the sum of the individual currents flowing through each of the loads connected in parallel.
- In a parallel combination of different bulbs, the bulb with the highest wattage glows with maximum brightness because of small resistance.
The power dissipated through a load is given by:
P = VI
Where V is the voltage across the load, and I is the current flowing through the load.
Calculation:
P = 60 W
V = 240 V
\(R=\frac{V^2}{P}=\frac{240^2}{60}=960\; Ω \)
The resistance of the bulb is 960 Ω.
