A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When R . . .

Question : A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is 

a) 230 Ω

b) 46 Ω

c) 26 Ω

d) 13 Ω

NEET Phase-II, 2017

Doubt by Manasvini

Solution : 

Case I : When the bulb is connected with 100 V source and it consume a power of 500 W.

We know, 
P=VI
Current passing through the Bulb (I)
I = P/V
I = 500/100
I = 5 A

Case II : When the bulb is connected with a resistance of R Ω in series with a source of 230 V. 

Here it is given that the bulb is still consuming the same power of 500 W. It means the current and voltage values across the bulb are still same. Since bulb and series are connected in series so we can say that current through them will be same and the sum of voltage drop across each of them will be equal to the total applied voltage.

Total applied voltage = Voltage across Bulb + Voltage across Resistance R

230 V = 100 V + V_R
V_R = 230-100
V_R = 130 V

Now, 
We know 
V=IR
V_R=I_RR
R=V_R/I_R
R=130/5
R=26Ω

Hence, c) 26 Ω, would be the correct option.