Three electric bulbs with same voltage ratings of 110 volts but wattage ratings of 40, 60 and 100 watts respectively are connected in series across a 220 volt supply line. If their brightness are , `B_(1), B_(2), B_(3)` respectively, then-

Options

A. `B_(1) gt B_(2) gt B_(3)`
B. `B_(1) gt B_(2) lt B_(3)`
C. `B_(1) = B_(2) = B_(3)`
D. Bulbs will burn out due to the high voltage supply

Explanation

To solve the problem, we need to determine the brightness of three electric bulbs connected in series across a 220-volt supply. The bulbs have the following specifications:

- Bulb 1: 40 watts
- Bulb 2: 60 watts
- Bulb 3: 100 watts

### Step 1: Calculate the resistance of each bulb

The power rating of a bulb is given by the formula:

\[ P = \frac{V^2}{R} \]

Where:
- \( P \) is the power (in watts)
- \( V \) is the voltage rating (in volts)
- \( R \) is the resistance (in ohms)

Given that all bulbs are rated for 110 volts, we can rearrange the formula to find the resistance:

\[ R = \frac{V^2}{P} \]

Now we can calculate the resistance for each bulb.

**For Bulb 1 (40 watts):**
\[ R_1 = \frac{110^2}{40} = \frac{12100}{40} = 302.5 \, \Omega \]

**For Bulb 2 (60 watts):**
\[ R_2 = \frac{110^2}{60} = \frac{12100}{60} \approx 201.67 \, \Omega \]

**For Bulb 3 (100 watts):**
\[ R_3 = \frac{110^2}{100} = \frac{12100}{100} = 121 \, \Omega \]

### Step 2: Analyze the resistance values

Now we have the resistances:
- \( R_1 = 302.5 \, \Omega \)
- \( R_2 \approx 201.67 \, \Omega \)
- \( R_3 = 121 \, \Omega \)

### Step 3: Determine brightness based on resistance

In a series circuit, the same current flows through each bulb. The brightness of a bulb is proportional to the power it dissipates, which can be expressed as:

\[ P = I^2 R \]

Since the current \( I \) is constant for all bulbs in series, the brightness is directly proportional to the resistance. Therefore, the bulb with the highest resistance will be the brightest.

### Step 4: Compare the resistances

From our calculations:
- \( R_1 > R_2 > R_3 \)

Thus, the brightness order is:
- \( B_1 > B_2 > B_3 \)

### Conclusion

The bulb with the highest resistance (Bulb 1, 40 watts) will be the brightest, followed by Bulb 2 (60 watts), and finally Bulb 3 (100 watts) will be the least bright.

### Final Answer

The order of brightness is:
\[ B_1 > B_2 > B_3 \]

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