Two electric bulbs rated at 25 W, 220 V and 100 W, 220 V are connected in series across a 220 V volt

Q: Why does the 25W bulb consume more power than the 100W bulb?

In a series circuit, the current flowing through all components is the same. The power consumed is given by $P = i^2 R$. Since the 25W bulb has a higher resistance than the 100W bulb, it dissipates more power and glows brighter.

Q: Can we use $P = V^2/R$ to find the consumed power?

You cannot use $P = V^2/R$ with $V = 220\text{ V}$ because the $220\text{ V}$ supply is divided between the two bulbs. The voltage across each individual bulb is less than $220\text{ V}$.

Q: What if the bulbs were connected in parallel?

If connected in parallel across $220\text{ V}$, each bulb would receive the full $220\text{ V}$. In that case, they would consume their rated powers: $25\text{ W}$ and $100\text{ W}$ respectively, and the $100\text{ W}$ bulb would glow brighter.

Q: Why didn't we calculate the exact decimal value of resistance initially?

Keeping the resistance as a fraction like $220^2/25$ allows for massive cancellations later when we calculate the current and power. It makes the math faster and prevents rounding errors.

Q: Does the total power consumed equal the sum of individual powers?

Yes! The total power consumed is $16\text{ W} + 4\text{ W} = 20\text{ W}$. You can also verify this by calculating $P_{\text{total}} = V_{\text{supply}}^2 / R_{\text{eq}}$.

Question

Two electric bulbs rated at $25	ext{ W}, 220	ext{ V}$ and $100	ext{ W}, 220	ext{ V}$ are connected in series across a $220	ext{ V}$ voltage source. If the $25	ext{ W}$ and $100	ext{ W}$ bulbs draw powers $P_1$ and $P_2$ respectively, then

Options:

(A) $P_1 = 16	ext{ W}, P_2 = 4	ext{ W}$

(B) $P_1 = 4	ext{ W}, P_2 = 16	ext{ W}$

(C) $P_1 = 9	ext{ W}, P_2 = 16	ext{ W}$

(D) $P_1 = 16	ext{ W}, P_2 = 9	ext{ W}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have two electric bulbs. One is rated at $25	ext{ W}, 220	ext{ V}$ and the other is rated at $100	ext{ W}, 220	ext{ V}$. You decide to connect them in series across a $220	ext{ V}$ power supply. A common misconception is that the $100	ext{ W}$ bulb will glow brighter because of its higher power rating. But wait, is that really true? Let's dive into the physics of it.

The power rating of a bulb tells us how much power it will consume *only* if it is connected across its rated voltage. Since these bulbs are connected in series, the $220	ext{ V}$ supply will be divided between them. Neither bulb will get the full $220	ext{ V}$. To find out what actually happens, we first need to determine the one property of the bulbs that remains constant: their resistance.

### The Master Equation

The resistance of any electrical appliance can be found using its rated values. The formula is:

$$R = \frac{V_{	ext{rated}}^2}{P_{	ext{rated}}}$$

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

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

For the second bulb ($100	ext{ W}$):

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

Notice that we are not calculating the exact numerical values just yet. Keeping them in this fractional form will make our upcoming calculations much smoother and elegant.

### Final Calculation

Since the bulbs are in series, the equivalent resistance of the circuit is simply the sum of their individual resistances, $R_{	ext{eq}} = R_1 + R_2$. The current $i$ flowing through the circuit is the same for both bulbs and is given by Ohm's law:

$$i = \frac{V_{	ext{supply}}}{R_{	ext{eq}}} = \frac{220}{\frac{220^2}{25} + \frac{220^2}{100}}$$

Here is where our strategy pays off. We can factor out $220^2$ from the denominator:

$$i = \frac{220}{220^2 \left(\frac{1}{25} + \frac{1}{100}ight)} = \frac{1}{220 \left(\frac{4 + 1}{100}ight)} = \frac{1}{220 \left(\frac{5}{100}ight)}$$

Simplifying this carefully, we find the current to be exactly:

$$i = \frac{1}{11} 	ext{ A}$$

Now, to find the actual power consumed by each bulb in this series circuit, we use the formula $P = i^2 R$. For the first bulb:

$$P_1 = i^2 R_1 = \left(\frac{1}{11}ight)^2 	imes \frac{220^2}{25} = \frac{1}{121} 	imes \frac{48400}{25} = 16 	ext{ W}$$

And for the second bulb:

$$P_2 = i^2 R_2 = \left(\frac{1}{11}ight)^2 	imes \frac{220^2}{100} = \frac{1}{121} 	imes \frac{48400}{100} = 4 	ext{ W}$$

**Conclusion:** The $25	ext{ W}$ bulb consumes $16	ext{ W}$ of power, while the $100	ext{ W}$ bulb consumes only $4	ext{ W}$. The bulb with the lower power rating actually glows brighter when connected in series! This beautiful paradox is a classic favorite in competitive exams.

Animated Solution for Physics - Current Electricity: Two electric bulbs rated at $25 W, 220 V$ and $100 W, 220 V$ are connected in series across a $220 V$ voltage source. If the $25 W$ and $100 W$ bulbs draw powers $P_{1}$ and $P_{2}$ respectively, then

Select Answer:

Visualized Solution

Circuit Setup

Resistance of a Bulb

Resistances $R_{1}$ and $R_{2}$

Current in Series Circuit

Calculating Current $i$

Simplifying Current $i$

Power Consumed

Power of Bulb 1 ( $P_{1}$ )

Power of Bulb 2 ( $P_{2}$ )

Conclusion

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

Analyzing the Setup

The Master Equation

Final Calculation

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