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

AIIMS 2008

Exam: AIIMS
Year: 2008
Subject: Physics
Chapter: Current Electricity

AIIMS 2008 · Physics · Current Electricity

A bulb of (25 ~W, 200 ~V ) and another bulb of (100 ~W, 200 ~V )...

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

Options

Aboth bulbs will glow with same brightness
Bboth bulbs will get fused
C(25 \mathrm{~W}) bulb will glow more brightly
D(100 \mathrm{~W}) bulb will glow more brightly.

Correct answer

C. (25 \mathrm{~W}) bulb will glow more brightly

Step-by-step 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.

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