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Question

Physics Question on Current electricity

A filament bulb (500W,100V)(500\, W, 100 \,V) is to be used in 230V230\, V main supply. When a resistance RR is connected in series, it works perfectly and the bulb consumes 500W500\, W. The value of RR is

A

230Ω230\, \Omega

B

46Ω46\, \Omega

C

26Ω26 \, \Omega

D

13Ω13 \, \Omega

Answer

26Ω26 \, \Omega

Explanation

Solution

P=V2RP = \frac{V^{2}}{R}
Rb=V2P=10000500=20ΩR_{b} = \frac{V^{2}}{P} = \frac{10000}{500} = 20\, \Omega
i=100Rb=130Ri = \frac{100}{R_{b}} = \frac{130}{R}
10020=130R\frac{100}{20} = \frac{130}{R}
R=26ΩR = 26\,\Omega