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An electrical bulb rated $220 \mathrm{~V}, 100 \mathrm{~W}$ , is connected in series with another bulb rated $220 \mathrm{~V}, 60 \mathrm{~W}$ . If the voltage across combination is $220 \mathrm{~V}$ , the power consumed by the $100 \mathrm{~W}$ bulb will be about____________ $\mathrm{~W}$
Option: 1
14

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Bulb 1

$\mathrm{(\left.V_R\right)_1=220 \mathrm{~V} }$ $\mathrm{\quad R_1=\frac{\left(V_R\right)_1^2}{\left(P_R\right)_1}=484 \Omega }$

$\mathrm{\left(P_R\right)_1=100 \mathrm{~W}}$

$\mathrm{\left ( V_{R1} \right )\&\left ( P_{R2} \right )}$ are voltage rating and power rating for the bulb1 respectively

Bulb 2
$\mathrm{\left(V_R\right)_2=220 \mathrm{~V} \\ }$ $\mathrm{\quad R_2=\frac{\left(V_R\right)_2^2}{\left(P_R\right)_2}=\frac{4840}{6} \Omega }$

$\mathrm{\left(P_R\right)_2=60 \mathrm{~W} }$

$\mathrm{220 =I\left(R_1+R_2\right) }$

$\mathrm{=I\left(484+\frac{4840}{6}\right) }$

$\mathrm{10 =I\left(22+\frac{220}{6}\right) }$

$\mathrm{I=\frac{60}{352} \mathrm{~A}}$
Power consumed by bulb 1

$\mathrm{P =I^2 R_1 }$

$\mathrm{=\frac{60 \times 60}{352 \times 352} \times(22 \times 22) }$

$\mathrm{=\frac{60 \times 60}{16 \times 16}=\frac{225}{16} }$

$\mathrm{P \cong 14 \mathrm{~W}}$

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shivangi.shekhar

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