Two electric bulbs marked 25 W-220 V and 100 W-220 V are connected in series to a 440 V supply. Whic

Q: Why does the 25W bulb fuse and not the 100W one?

In a series circuit, the voltage divides in proportion to the resistance. The 25W bulb has a much higher resistance ($1936\ \Omega$) compared to the 100W bulb ($484\ \Omega$). Therefore, it takes a much larger share of the 440V supply ($352\text{ V}$), which exceeds its 220V rating, causing it to fuse.

Q: Can we just add the power ratings in series?

No! Power ratings are only valid when the bulb is operated exactly at its rated voltage. In a series circuit, the actual voltage across each bulb changes, so the actual power consumed is completely different from the rated power.

Q: What happens to the 100W bulb after the 25W bulb fuses?

Once the 25W bulb fuses, its filament breaks. This creates an open circuit, stopping the flow of current entirely. Consequently, the 100W bulb will immediately turn off and remain safe.

Q: Why do we use $R = V^2/P$ instead of $P = I^2 R$ to find resistance?

The rated voltage and rated power are fixed, physical properties printed on the bulb by the manufacturer. The current $I$, however, depends entirely on how you connect the bulb in a circuit. Therefore, we must use the fixed rated values to find the constant resistance.

Q: Would the 100W bulb fuse if connected alone to the 440V supply?

Yes, absolutely! If connected alone, the entire 440V would drop across the 100W bulb. Since 440V is double its rated 220V, the filament would overheat and fuse instantly.

Question

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

Options:

(A) Both

(B) 100 W

(C) 25 W

(D) None of these

Sigma Percentile · Accepted Answer

The Tale of Two Bulbs: A Fiery Series Circuit

Imagine you are setting up the lighting for a grand event. You have two electric bulbs in your hands. One is a modest $25	ext{ W}$ bulb, and the other is a bright $100	ext{ W}$ bulb. Both are designed to operate safely at a standard $220	ext{ V}$.

But here is the twist: you decide to connect them in **series** across a massive $440	ext{ V}$ power supply! What happens next? Does the extra voltage distribute evenly, or is one of the bulbs destined for a fiery end? Let's dive into the physics of this explosive scenario.

### Uncovering the Hidden Resistance

To understand how these bulbs will behave, we first need to look at their internal anatomy—specifically, the resistance of their tungsten filaments. The resistance is a fixed property of the bulb, determined by its manufacturing. We can find it using the power formula:

$$P = \frac{V^2}{R} \implies R = \frac{V^2}{P}$$

Let's calculate the resistance for both bulbs. For the $25	ext{ W}$ bulb:

$$R_1 = \frac{220^2}{25} = \frac{48400}{25} = 1936\ \Omega$$

And for the $100	ext{ W}$ bulb:

$$R_2 = \frac{220^2}{100} = \frac{48400}{100} = 484\ \Omega$$

Notice something fascinating here? The bulb with the **lower power rating** has a **much higher resistance**. This is a crucial insight! The $25	ext{ W}$ bulb is like a narrow, restrictive pipe, while the $100	ext{ W}$ bulb is a wider, easier path for the current.

### The Flow of Current

Since the bulbs are connected in series, the total resistance of the circuit is simply the sum of their individual resistances:

$$R_{	ext{eq}} = R_1 + R_2 = 1936 + 484 = 2420\ \Omega$$

Now, we can use Ohm's Law to find the current flowing through this series circuit. The total supply voltage is $440	ext{ V}$:

$$I = \frac{V_{	ext{supply}}}{R_{	ext{eq}}} = \frac{440}{2420} = \frac{2}{11}	ext{ A}$$

Because they are in series, this exact same current of $\frac{2}{11}	ext{ A}$ flows through both bulbs. They are forced to share the same flow, regardless of their different resistances.

### The Moment of Truth: Voltage Drops

Here is where the drama unfolds. In a series circuit, the total voltage is divided among the components based on their resistance. The component with the higher resistance takes a larger share of the voltage. Let's calculate the exact voltage drop across each bulb using $V = I 	imes R$.

For the $25	ext{ W}$ bulb:

$$V_1 = I 	imes R_1 = \frac{2}{11} 	imes 1936 = 352	ext{ V}$$

For the $100	ext{ W}$ bulb:

$$V_2 = I 	imes R_2 = \frac{2}{11} 	imes 484 = 88	ext{ V}$$

### The Verdict

Look closely at those numbers! The $25	ext{ W}$ bulb, which is only rated to handle $220	ext{ V}$, is being subjected to a crushing $352	ext{ V}$. The filament will overheat, melt, and **fuse** almost instantly.

On the other hand, the $100	ext{ W}$ bulb is only experiencing $88	ext{ V}$, which is well below its $220	ext{ V}$ limit. It would survive (though it would glow very dimly if the circuit hadn't broken).

This is a classic trap in physics: we often assume the higher power bulb is the "stronger" one. But in a series circuit, the lower power bulb has the higher resistance, takes the lion's share of the voltage, and is the first to blow!

Animated Solution for Physics - Current Electricity: Two electric bulbs marked 25 W-220 V and 100 W-220 V are connected in series to a 440 V supply. Which of the bulbs will fuse?

Select Answer:

Visualized Solution

Circuit Setup

Resistance Formula

Calculating Resistances

Circuit Current

Voltage Drops

Conclusion

The Way Forward

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Uncovering the Hidden Resistance

The Flow of Current

The Moment of Truth

The Verdict

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