Three bulbs of 40 W , 60 W , 100 W are arranged in series with 220 volt supply which bulb has minimum resistance

Options

A. 100 W
B. 40W
C. 60W
D. Equal in all bulbs

Answer

To determine which bulb has the minimum resistance among the three bulbs (40 W, 60 W, and 100 W) connected in series to a 220 V supply, we can follow these steps:

### Step 1: Understand the relationship between power, voltage, and resistance
The power (P) consumed by an electrical device is related to the voltage (V) across it and its resistance (R) by the formula:
\[ P = \frac{V^2}{R} \]
From this, we can rearrange the formula to find the resistance:
\[ R = \frac{V^2}{P} \]

### Step 2: Calculate the resistance of each bulb
Since we know the power ratings of the bulbs, we can calculate the resistance for each bulb using the formula derived above. However, we need to find the voltage across each bulb when they are connected in series.

In a series circuit, the same current flows through each component. The voltage across each bulb can be calculated if we know the total voltage and the power ratings.

### Step 3: Calculate the equivalent resistance for each bulb
1. **For the 40 W bulb:**
\[ R_{40} = \frac{V^2}{P} = \frac{(220)^2}{40} = \frac{48400}{40} = 1210 \, \Omega \]

2. **For the 60 W bulb:**
\[ R_{60} = \frac{V^2}{P} = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega \]

3. **For the 100 W bulb:**
\[ R_{100} = \frac{V^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 4: Compare the resistances
Now we can compare the resistances calculated:
- Resistance of 40 W bulb: \( 1210 \, \Omega \)
- Resistance of 60 W bulb: \( 806.67 \, \Omega \)
- Resistance of 100 W bulb: \( 484 \, \Omega \)

### Step 5: Identify the bulb with minimum resistance
From the calculated resistances, we can see that the 100 W bulb has the lowest resistance of \( 484 \, \Omega \).

### Conclusion
Thus, the bulb with the minimum resistance is the **100 W bulb**.

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