A bulb of 25 ~W, 200 ~V and another bulb of 100 ~W, 200 ~V are connected in series with a supply line of 220 ~V . Then

Correct answer: (C) 25 ~W bulb will glow more brightly. Resistance of 25 ~W, 200 ~V bulb R_1=((200)^2)/25 Ω Resistance of 100 ~W, 200 ~V bulb R_2=((200)^2)/100 Ω . Clearly, R_1>R_2 . ∴ H_1>H_2 . ∴ 25 ~W bulb will glow more brightly.

Q108 / 180 · AIIMS 2008 · Physics

A bulb of $25 \mathrm{~W}, 200 \mathrm{~V}$ and another bulb of $100 \mathrm{~W}, 200 \mathrm{~V}$ are connected in series with a supply line of $220 \mathrm{~V}$. Then

A
both bulbs will glow with same brightness
B
both bulbs will get fused
Correct answer
C
$25 \mathrm{~W}$ bulb will glow more brightly
D
$100 \mathrm{~W}$ bulb will glow more brightly.

Solution

Resistance of $25 \mathrm{~W}, 200 \mathrm{~V}$ bulb $R_1=\frac{(200)^2}{25} \Omega$ Resistance of $100 \mathrm{~W}, 200 \mathrm{~V}$ bulb $R_2=\frac{(200)^2}{100} \Omega$. Clearly, $R_1>R_2 . \quad \therefore \quad H_1>H_2$. $\therefore 25 \mathrm{~W}$ bulb will glow more brightly.

Hence, the correct answer is option: 'C'.

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A bulb of \(25 \mathrm{~W}, 200 \mathrm{~V}\) and another bulb of \(100 \mathrm{~W}, 200 \mathrm{~V}\) are connected in series with a supply line of \(220 \mathrm{~V}\). Then

Solution

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