Three 100 W light bulbs are connected in series to a 220V power source. If two of the light bulbs are replaced by 60 W bulbs, the brightness of the remaining on w bulb is

Options

A. Brighter than it was before
B. Dinner than it was before
C. The same brightness as it was before
D. will not illuminate

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Calculate the total resistance of the original circuit
In the original circuit, there are three 100 W light bulbs connected in series.

First, we need to find the resistance of each bulb. The power (P) of a bulb is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the bulb and \( R \) is its resistance.

For a 100 W bulb connected to a 220 V source, we can rearrange the formula to find the resistance:
\[ R = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

Since there are three bulbs in series, the total resistance \( R_{total} \) is:
\[ R_{total} = R_1 + R_2 + R_3 = 484 + 484 + 484 = 1452 \, \Omega \]

### Step 2: Calculate the current flowing through the circuit
Using Ohm's law, the total current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{V_{total}}{R_{total}} = \frac{220}{1452} \approx 0.151 \, A \]

### Step 3: Calculate the voltage drop across each bulb in the original circuit
Since the bulbs are identical, the voltage drop across each bulb \( V_b \) is:
\[ V_b = I \times R = 0.151 \times 484 \approx 73.33 \, V \]

### Step 4: Analyze the new circuit with two 60 W bulbs and one 100 W bulb
Now, we replace two of the 100 W bulbs with 60 W bulbs. We need to calculate the resistance of the new bulbs.

For a 60 W bulb:
\[ R_{60} = \frac{220^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega \]

Now, the total resistance of the new circuit with two 60 W bulbs and one 100 W bulb is:
\[ R_{total\_new} = R_{60} + R_{60} + R_{100} = 806.67 + 806.67 + 484 \approx 2097.34 \, \Omega \]

### Step 5: Calculate the new current flowing through the circuit
Using Ohm's law again, the new total current \( I_{new} \) is:
\[ I_{new} = \frac{V_{total}}{R_{total\_new}} = \frac{220}{2097.34} \approx 0.105 \, A \]

### Step 6: Calculate the voltage drop across the 100 W bulb in the new circuit
The voltage drop across the 100 W bulb \( V_{100} \) is:
\[ V_{100} = I_{new} \times R_{100} = 0.105 \times 484 \approx 50.7 \, V \]

### Step 7: Determine the brightness of the 100 W bulb
The brightness of a bulb is proportional to the power it dissipates, which can be calculated as:
\[ P_{100} = I_{new}^2 \times R_{100} = (0.105)^2 \times 484 \approx 5.4 \, W \]

### Conclusion
The brightness of the remaining 100 W bulb will be significantly lower than before, as it is now dissipating only about 5.4 W.