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A filament bulb (500 W, 100 V) is to be ...

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 consumers `500 W`. The value of `R` is

A

`230 Omega`

B

`46 Omega`

C

`25 Omega`

D

`13 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
If a rated voltage and power are given,
then `P_("rated") = (V_("rated")^(2))/R`
`therefore` Current in the bulb, `I=P/V` `[therefore`P=VI]
`I= 500/100= 5A`
`therefore` Resistance of bulb, `R_(b) = (100 xx 100)/(500) = 20 Omega`
`therefore` Resistance R is connected in series.
`therefore` Current, `I= E/R_("net") = 230/(R+R_(b))`
`therefore rArr R+20 = 230/5 =46`
`R= 26Omega`
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