Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440V supply. Which of the bulbs will fuse?

Current Electricity PYQ — JEE Main 2012 · Shift 0 · Q58 with Solution

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Correct answer: C

Find the resistance of each bulb from its rating

For a bulb rated $P$ at voltage $V$ , its resistance at rated conditions is

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

For the $25\text{ W}, 220\text{ V}$ bulb:

R_{25} = \frac{220^2}{25} = \frac{48400}{25} = 1936\,\Omega.

For the $100\text{ W}, 220\text{ V}$ bulb:

R_{100} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega.

Since they are in series, find the current through the circuit

Total resistance:

R_{\text{total}} = 1936 + 484 = 2420\,\Omega.

Applied voltage is $440\text{ V}$ , so current is

I = \frac{440}{2420} = \frac{2}{11}\text{ A} \approx 0.182\text{ A}.

Find the voltage across each bulb

Across the $25\text{ W}$ bulb:

V_{25} = I R_{25} = \frac{2}{11}\times 1936 = 352\text{ V}.

Across the $100\text{ W}$ bulb:

V_{100} = I R_{100} = \frac{2}{11}\times 484 = 88\text{ V}.

Compare with rated voltage

Each bulb is rated for $220\text{ V}$ .

The $25\text{ W}$ bulb gets $352\text{ V}$ , which is much greater than $220\text{ V}$ , so it will fuse.
The $100\text{ W}$ bulb gets only $88\text{ V}$ , so it will not fuse.

Check by power dissipated

Power in each bulb in series:

P = I^2R.

For $25\text{ W}$ bulb:

P_{25} = \left(\frac{2}{11}\right)^2 \times 1936 = 64\text{ W}.

This is greater than its rated $25\text{ W}$ , so it fuses.

For $100\text{ W}$ bulb:

P_{100} = \left(\frac{2}{11}\right)^2 \times 484 = 16\text{ W}.

This is much less than its rated $100\text{ W}$ , so it is safe.

Conclusion

The bulb that will fuse is the $25\text{ W}$ bulb.

So, the correct option is C.

More from Current Electricity

View written solutionFree

Correct answer: C

Find the resistance of each bulb from its rating

For a bulb rated $P$ at voltage $V$ , its resistance at rated conditions is

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

For the $25\text{ W}, 220\text{ V}$ bulb:

R_{25} = \frac{220^2}{25} = \frac{48400}{25} = 1936\,\Omega.

For the $100\text{ W}, 220\text{ V}$ bulb:

R_{100} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega.

Since they are in series, find the current through the circuit

Total resistance:

R_{\text{total}} = 1936 + 484 = 2420\,\Omega.

Applied voltage is $440\text{ V}$ , so current is

I = \frac{440}{2420} = \frac{2}{11}\text{ A} \approx 0.182\text{ A}.

Find the voltage across each bulb

Across the $25\text{ W}$ bulb:

V_{25} = I R_{25} = \frac{2}{11}\times 1936 = 352\text{ V}.

Across the $100\text{ W}$ bulb:

V_{100} = I R_{100} = \frac{2}{11}\times 484 = 88\text{ V}.

Compare with rated voltage

Each bulb is rated for $220\text{ V}$ .

The $25\text{ W}$ bulb gets $352\text{ V}$ , which is much greater than $220\text{ V}$ , so it will fuse.
The $100\text{ W}$ bulb gets only $88\text{ V}$ , so it will not fuse.

Check by power dissipated

Power in each bulb in series:

P = I^2R.

For $25\text{ W}$ bulb:

P_{25} = \left(\frac{2}{11}\right)^2 \times 1936 = 64\text{ W}.

This is greater than its rated $25\text{ W}$ , so it fuses.

For $100\text{ W}$ bulb:

P_{100} = \left(\frac{2}{11}\right)^2 \times 484 = 16\text{ W}.

This is much less than its rated $100\text{ W}$ , so it is safe.

Conclusion

The bulb that will fuse is the $25\text{ W}$ bulb.

So, the correct option is C.

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