Correct Answer - Option 1 : 2130 Ω
Concept:
The intensity of a bulb is directly proportional to the power dissipated by the bulb.
The power dissipated is given by
\(P = \frac{{{V^2}}}{R}\)
Where V is the voltage applied across the bulb
R is the resistance of the lamp
Calculation:
The power rating of lamp = 25 W
Voltage rating of lamp = 230 V
Resistance \(R = \frac{{{V^2}}}{P} = \frac{{{{230}^2}}}{{25}} =2116\;{\rm{\Omega }}\)
Resistance = 2130 Ω (approx)
