Two bulbs one of 200 volts , 60 watts & the other of 200 volts , 100 watts are connected in series to a 200 volt supply . The power consumed will be Assume the light bulbs are rated for single connection to 120 V. With the mistaken connection , the power dissipated by each bulb is

Options

A. 37 .5 watt
B. 160 watt
C. 62 .5 watt
D. 100 watt

Answer

To solve the problem, we need to determine the power dissipated by each bulb when they are connected in series to a 200V supply. We have two bulbs: one rated at 200V and 60W (Bulb 1) and the other rated at 200V and 100W (Bulb 2).

### Step-by-Step Solution:

1. **Calculate the Resistance of Each Bulb:**
The resistance of a bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
- For Bulb 1 (60W):
\[ R_1 = \frac{200^2}{60} = \frac{40000}{60} \approx 666.67 \, \Omega \]
- For Bulb 2 (100W):
\[ R_2 = \frac{200^2}{100} = \frac{40000}{100} = 400 \, \Omega \]

2. **Calculate the Total Resistance in Series:**
The total resistance \( R_{total} \) for resistors in series is simply the sum of their resistances:
\[ R_{total} = R_1 + R_2 = 666.67 + 400 = 1066.67 \, \Omega \]

3. **Calculate the Current through the Circuit:**
Using Ohm's law, the current \( I \) can be calculated as:
\[ I = \frac{V}{R_{total}} = \frac{200}{1066.67} \approx 0.187 \, A \]

4. **Calculate the Power Dissipated by Each Bulb:**
The power dissipated by each bulb can be calculated using the formula:
\[ P = I^2 R \]
- For Bulb 1:
\[ P_1 = I^2 R_1 = (0.187)^2 \times 666.67 \approx 23.43 \, W \]
- For Bulb 2:
\[ P_2 = I^2 R_2 = (0.187)^2 \times 400 \approx 14.06 \, W \]

5. **Final Result:**
The power dissipated by Bulb 1 is approximately 23.43W, and the power dissipated by Bulb 2 is approximately 14.06W.

### Summary:
- Power dissipated by Bulb 1: **23.43 W**
- Power dissipated by Bulb 2: **14.06 W**