Test Index

VITEEE 2010 Paper

Question : 28 of 120

Marks: +1, -0

Solution:

Power,

P = \frac{V^{2}}{R},

\Rightarrow R = \frac{V^{2}}{P}

For the first bulb,

R_{1}=\left(\frac{V^{2}}{P_{1}}\right)=\left(\frac{(220)^{2}}{25}\right)=1936\,\Omega

For the second bulb,

R_{2}=\left(\frac{V^{2}}{P_{2}}\right)=\left(\frac{(220)^{2}}{100}\right)=48\,\Omega

Current in series combination is the same in the two bulbs,

i=\frac{V}{R_{1}+R_{2}}=\frac{220}{1936+484}=\frac{220}{2420}=\frac{1}{11}\,\text{A}

If the actual powers in the two bulbs be

P_{1}

and

P_{2}

then

P'_{1}=i^{2}R_{1}=\left(\frac{1}{11}\right)^{2}\times 1936=16\,\text{W}

P'_{2}=i^{2}R_{2}=\left(\frac{1}{11}\right)^{2}\times 484=4\,\text{W}

Since

P'_{1}>P'_{2},

so, 25 W bulb will glow more brightly.

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