Two bulbs one of `200` volts, `60` watts & the other of `200` volts, `100` watts are connected in series to a `200` volts supply. The power consumed will be

Options

A. `37.5` watt
B. `160` watt
C. `62.5` watt
D. `110` watt

Explanation

To solve the problem, we need to find the total power consumed by the two bulbs connected in series to a 200 volts supply. Here’s a step-by-step solution:

### 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 of the bulb and \( P \) is the power rating.

1. **For the first bulb (200V, 60W)**:
\[ R_1 = \frac{200^2}{60} = \frac{40000}{60} = \frac{2000}{3} \approx 666.67 \, \Omega \]

2. **For the second bulb (200V, 100W)**:
\[ R_2 = \frac{200^2}{100} = \frac{40000}{100} = 400 \, \Omega \]

### Step 2: Calculate the Total Resistance in Series

When resistors are connected in series, the total resistance \( R_{total} \) is the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 \]
Substituting the values we calculated:
\[ R_{total} = \frac{2000}{3} + 400 \]
To add these, convert \( 400 \) to a fraction:
\[ 400 = \frac{1200}{3} \]
Now add:
\[ R_{total} = \frac{2000}{3} + \frac{1200}{3} = \frac{3200}{3} \approx 1066.67 \, \Omega \]

### Step 3: Calculate the Total Power Consumed

The power consumed by the series circuit can be calculated using the formula:
\[ P = \frac{V^2}{R_{total}} \]
Substituting the values:
\[ P = \frac{200^2}{\frac{3200}{3}} = \frac{40000}{\frac{3200}{3}} = 40000 \times \frac{3}{3200} = \frac{120000}{3200} = 37.5 \, W \]

### Final Answer

The total power consumed by the two bulbs connected in series to a 200 volts supply is:
\[ \boxed{37.5 \, W} \]

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