A filament bulb (500 ,W, 100 ,V) is to be use | vedclass

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(A) First,calculate the resistance of the bulb $(R_B)$:$R_B = \frac{V^2}{P} = \frac{(100)^2}{500} = \frac{10000}{500} = 20 \, \Omega$.Next,calculate t

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(A) First,calculate the resistance of the bulb $(R_B)$:$R_B = \frac{V^2}{P} = \frac{(100)^2}{500} = \frac{10000}{500} = 20 \, \Omega$.Next,calculate the current $(I)$ flowing through the bulb when it operates at its rated power:$I = \frac{P}{V} = \frac{500}{100} = 5 \, A$.Since the bulb and resistor $R$ are in series,the same current $I = 5 \, A$ flows through the resistor $R$. The voltage drop across the resistor $(V_R)$ is the difference between the supply voltage and the bulb's rated voltage:$V_R = V_{supply} - V_{bulb} = 230 \, V - 100 \, V = 130 \, V$.Using Ohm's law for the resistor $R$:$V_R = I 	imes R$$130 = 5 	imes R$$R = \frac{130}{5} = 26 \, \Omega$.

$A$ 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 .................. $\Omega$.

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