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VITEEE 2015 Paper

Section: Physics

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Question : 16 of 120

Marks: +1, -0

A 30V -90 W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is

10 Ω
30 Ω
20 Ω
40 Ω

Solution:

Resistance of lamp

R

=

\frac{C^2}{P}

=

\frac{(30)^2}{90}

= 10 Ω

Current in the lamp

I =

\frac{V}{R_0}

=

\frac{30}{10}

= 3 A

As the lamp is operated on 120V DC, then resistance becomes

R' =

\frac{V'}{i}

=

\frac{120}{3}

= 40 Ω

For proper glow, a resistance R is joined in series with the bulb

R' = R +

R_0

⇒

R^{\alpha}

= R' -

R_0

= 40 - 10 = 30 Ω

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