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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` isA. `230 Omega`B. `46 Omega`C. `25 Omega`D. `13 Omega` |
Answer» Correct Answer - 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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