Two bulbs 60 W and 100 W designed for voltage 220 V are connected in series across 220 V source. The net power dissipated is

Options

A. 80 W
B. 160 W
C. `37.5 W`
D. 60 W

Answer

To solve the problem of finding the net power dissipated when two bulbs (60 W and 100 W) designed for 220 V are connected in series across a 220 V source, follow these steps:

### Step 1: Calculate the resistance of each bulb
The resistance \( R \) of each bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage rating (220 V) and \( P \) is the power rating of the bulb.

**For the 60 W bulb (B1)**:
\[ R_1 = \frac{220^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega \]

**For the 100 W bulb (B2)**:
\[ R_2 = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 2: Calculate the equivalent resistance of the series connection
When resistors are connected in series, the total resistance \( R_{eq} \) is the sum of the individual resistances:
\[ R_{eq} = R_1 + R_2 \]
Substituting the values calculated:
\[ R_{eq} = 806.67 + 484 = 1290.67 \, \Omega \]

### Step 3: Calculate the total power dissipated in the circuit
The total power \( P_{total} \) dissipated in a circuit can be calculated using the formula:
\[ P_{total} = \frac{V^2}{R_{eq}} \]
Substituting the values:
\[ P_{total} = \frac{220^2}{1290.67} = \frac{48400}{1290.67} \approx 37.5 \, W \]

### Final Answer
The net power dissipated by the two bulbs when connected in series across a 220 V source is approximately **37.5 W**.

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