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There area $25 W-220 \mathrm{~V}$ bulb and a $100 \mathrm{~W}-220 \mathrm{~V}$ line. Which eletric bulb will glow more brightly?
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Verified Answer
The correct answer is:
$25 \mathrm{~W}$ bulb
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,
$$
R_{2}=\left(\frac{V^{2}}{P_{2}}\right)=\left(\frac{(220)^{2}}{100}\right)=484 \Omega
$$
Current in series combination is the same in the two bulbs,
$i=\frac{V}{R_{1}+R_{2}}=\frac{220}{1936+484}=\frac{220}{2420}=\frac{1}{11} A$
If the actual powers in the two bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{\prime}=i^{2} R_{1}=\left(\frac{1}{11}\right)^{2} \times 1936=16 \mathrm{~W}$
and $P_{2}^{\prime}=i^{2} R_{2}=\left(\frac{1}{11}\right)^{2} \times 484=4 \mathrm{~W}$
Since $P_{1}^{\prime}>P_{2}^{\prime}$, so, $25 \mathrm{~W}$ bulb will glow more brightly.
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,
$$
R_{2}=\left(\frac{V^{2}}{P_{2}}\right)=\left(\frac{(220)^{2}}{100}\right)=484 \Omega
$$
Current in series combination is the same in the two bulbs,
$i=\frac{V}{R_{1}+R_{2}}=\frac{220}{1936+484}=\frac{220}{2420}=\frac{1}{11} A$
If the actual powers in the two bulbs be $P_{1}$ and $P_{2}$ then
$P_{1}^{\prime}=i^{2} R_{1}=\left(\frac{1}{11}\right)^{2} \times 1936=16 \mathrm{~W}$
and $P_{2}^{\prime}=i^{2} R_{2}=\left(\frac{1}{11}\right)^{2} \times 484=4 \mathrm{~W}$
Since $P_{1}^{\prime}>P_{2}^{\prime}$, so, $25 \mathrm{~W}$ bulb will glow more brightly.
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