Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases w

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

In household circuits, all appliances (like these bulbs) are connected in parallel to the main supply. This means they all receive the same constant voltage $V$. Using $P = V^2/R$ allows us to easily compare their resistances because $V$ is constant.

Q: Does the resistance really change that much from room temperature to operating temperature?

Yes! Tungsten has a high positive temperature coefficient of resistance. When a bulb is turned on and the filament becomes white-hot, its resistance can be 10 to 15 times higher than its cold resistance at room temperature.

Q: If the resistance changes, how can we use the rated power to find the room temperature resistance?

The rated power gives us the *hot* resistance. However, assuming the bulbs are made of the same tungsten material, the ratio of their hot resistance to cold resistance is roughly the same. Therefore, the proportional relationships we derive for the hot state also hold true for the cold state.

Q: Why are there two correct options for this question?

This was a slightly flawed question in the exam. Because $100 > 60 > 40$, option (d) is correct. But also, because $100 = 60 + 40$, option (a) is exactly mathematically correct as well. Both are valid conclusions!

Q: Which bulb has the thickest filament?

The $100\text{ W}$ bulb has the lowest resistance. Since resistance is inversely proportional to the cross-sectional area ($R = \rho L / A$), the bulb with the lowest resistance must have the thickest filament (assuming their lengths are similar).

Question

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100	ext{ W}$, $60	ext{ W}$ and $40	ext{ W}$ bulbs have filament resistances $R_{100}$, $R_{60}$ and $R_{40}$, respectively, the relation between these resistances is

Options:

(A) $\frac{1}{R_{100}} = \frac{1}{R_{40}} + \frac{1}{R_{60}}$

(B) $R_{100} = R_{40} + R_{60}$

(C) $R_{100} > R_{60} > R_{40}$

(D) $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$

Sigma Percentile · Accepted Answer

### The Everyday Physics of Light Bulbs
Have you ever looked at the bulbs in your house and wondered what makes a $100	ext{ W}$ bulb shine brighter than a $40	ext{ W}$ bulb? It all comes down to the tiny tungsten filament inside and its electrical resistance.

In this problem, we are given three bulbs rated at $100	ext{ W}$, $60	ext{ W}$, and $40	ext{ W}$. Our goal is to find the mathematical relationship between their resistances at room temperature.

### The Master Equation
To connect power and resistance, we need to recall the formula for electrical power. Since all household appliances are connected in parallel, they all receive the same standard voltage $V$ (usually $220	ext{ V}$ in India).

Therefore, the most useful form of the power equation is:
$$P = \frac{V^2}{R}$$

Rearranging this to solve for the reciprocal of resistance, we get:
$$\frac{1}{R} = \frac{P}{V^2}$$

Since $V^2$ is a constant for all three bulbs, we can clearly see that the reciprocal of the resistance is directly proportional to the power rating:
$$\frac{1}{R} \propto P$$

### Analyzing the Inequalities
Let's apply this proportionality to our three bulbs. We have:
$$\frac{1}{R_{100}} = \frac{100}{V^2}$$
$$\frac{1}{R_{60}} = \frac{60}{V^2}$$
$$\frac{1}{R_{40}} = \frac{40}{V^2}$$

Since $100 > 60 > 40$, it immediately follows that:
$$\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$$

This perfectly matches one of the given options! But is that the whole story?

### The Hidden Additive Relationship
If we look closely at the power ratings, we can spot a simple arithmetic truth:
$$100 = 60 + 40$$

Because $\frac{1}{R} = \frac{P}{V^2}$, we can divide the entire equation by $V^2$:
$$\frac{100}{V^2} = \frac{60}{V^2} + \frac{40}{V^2}$$

Substituting the resistance terms back in, we arrive at a beautiful exact relationship:
$$\frac{1}{R_{100}} = \frac{1}{R_{60}} + \frac{1}{R_{40}}$$

This means that **both option (a) and option (d) are mathematically correct!** This was a slightly flawed question in the exam, but it serves as a brilliant conceptual test.

### The "Room Temperature" Nuance
You might be wondering why the question specifically mentioned "room temperature". The power rating of a bulb is defined at its operating temperature, where the tungsten filament is white-hot.

Because tungsten has a positive temperature coefficient of resistance, its resistance increases significantly as it heats up. However, the ratio of the hot resistance to the cold (room temperature) resistance is roughly the same for bulbs made of the same material. Therefore, the proportional relationships we derived hold true whether the bulbs are turned off or glowing brightly!

Select Answer:

Visualized Solution

Visualizing the Bulbs

The Power-Resistance Relation

Setting up the Proportions

Comparing the Values

The Hidden Additive Relation

The Temperature Nuance

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

The Everyday Physics of Light Bulbs

The Master Equation To connect power and resistance, we need to recall the formula for electrical power

Analyzing the Inequalities Let's apply this proportionality to our three bulbs

The Hidden Additive Relationship

The "Room Temperature" Nuance You might be wondering why the question specifically mentioned "room temperature"

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