Looking for classes? Ksquare Career Institute, Bengaluru →
Series resistance and electrical power
Concepts tested here
- series-resistor-bulb
All Questions
2016 Phase II 1 question
-
A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is:
Same current, so voltage divides in the ratio of resistances.
$R_{bulb} = \frac{V^2}{P} = \frac{100^2}{500} = 20\ \Omega$
R drops the remaining 130 V, so $\frac{R}{R_{bulb}} = \frac{130}{100}$
$\frac{R}{20} = \frac{130}{100} \Rightarrow R = 26\ \Omega$