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 find the net power dissipated by the two bulbs (60 W and 100 W) connected in series across a 220 V source, we can follow these steps:

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

- For the 60 W bulb:
\[ R_1 = \frac{220^2}{60} \]

- For the 100 W bulb:
\[ R_2 = \frac{220^2}{100} \]

### Step 2: Calculate the Equivalent Resistance
Since the bulbs are connected in series, the total or equivalent resistance \( R_{eq} \) is the sum of the individual resistances:
\[ R_{eq} = R_1 + R_2 \]

### Step 3: Substitute the Values
Substituting the values of \( R_1 \) and \( R_2 \) into the equation for \( R_{eq} \):
\[ R_{eq} = \frac{220^2}{60} + \frac{220^2}{100} \]
Factoring out \( 220^2 \):
\[ R_{eq} = 220^2 \left( \frac{1}{60} + \frac{1}{100} \right) \]

### Step 4: Simplify the Expression
To simplify \( \frac{1}{60} + \frac{1}{100} \):
\[ \frac{1}{60} + \frac{1}{100} = \frac{100 + 60}{6000} = \frac{160}{6000} \]
Thus,
\[ R_{eq} = 220^2 \cdot \frac{160}{6000} \]

### Step 5: Calculate the Net Power Dissipated
The net power \( P_{net} \) dissipated in the circuit can be calculated using the formula:
\[ P_{net} = \frac{V^2}{R_{eq}} \]
Substituting \( R_{eq} \):
\[ P_{net} = \frac{220^2}{R_{eq}} = \frac{220^2}{220^2 \cdot \frac{160}{6000}} = \frac{6000}{160} \]

### Step 6: Final Calculation
Calculating the final value:
\[ P_{net} = \frac{6000}{160} = 37.5 \text{ W} \]

Thus, the net power dissipated by the two bulbs is **37.5 W**.

---