Question

A bulb is marked with 60 W/240 V. The resistance when lighted at 240 V is -
1. 240 Ω
2. 480 Ω
3. 560 Ω
4. 960 Ω

Answer 1

Correct Answer - Option 4 : 960 Ω

Concept:

Electric Power: 

• The rate of consumption of electrical energy is electric power. 
• The SI unit of power is Watt.
• The power dissipated by an electrical device at potential difference across it V, and resistance R is given as

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

P is power, V is the potential difference, R is resistance

Calculation:

Given Power P = 60 W

Potential difference V = 240 Volt. 

Resistance R = ?

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

\(\implies R = \frac{240^2}{60} \Omega\)

⇒ R = 960 Ω

So, the correct option is 960 Ω.