When three identical bulbs of `60 W, 200 V` rating are connected in series to a `200 V` supply, the power drawn by them will be

Options

A. 60 W
B. 180 W
C. 10 W
D. 20 W

Answer

To solve the problem of finding the power drawn by three identical bulbs of 60 W, 200 V rating when connected in series to a 200 V supply, we can follow these steps:

### Step 1: Calculate the resistance of one bulb
The power (P) of a bulb is given by the formula:
\[ P = \frac{V^2}{R} \]
where V is the voltage rating of the bulb and R is its resistance.

Rearranging the formula to find R, we have:
\[ R = \frac{V^2}{P} \]

Substituting the values for one bulb:
\[ R = \frac{(200 \, \text{V})^2}{60 \, \text{W}} \]
\[ R = \frac{40000}{60} \]
\[ R = \frac{4000}{6} \]
\[ R = \frac{2000}{3} \, \Omega \]

### Step 2: Calculate the total resistance when bulbs are connected in series
When resistors (or bulbs in this case) are connected in series, the total resistance (R_total) is the sum of the individual resistances:
\[ R_{\text{total}} = R + R + R = 3R \]

Substituting the value of R we calculated:
\[ R_{\text{total}} = 3 \times \frac{2000}{3} \]
\[ R_{\text{total}} = 2000 \, \Omega \]

### Step 3: Calculate the total power drawn by the bulbs
The total power (P_total) drawn by the bulbs when connected to a voltage supply can be calculated using the formula:
\[ P_{\text{total}} = \frac{V^2}{R_{\text{total}}} \]

Substituting the values:
\[ P_{\text{total}} = \frac{(200 \, \text{V})^2}{2000 \, \Omega} \]
\[ P_{\text{total}} = \frac{40000}{2000} \]
\[ P_{\text{total}} = 20 \, \text{W} \]

### Final Answer
The power drawn by the three bulbs when connected in series to a 200 V supply is **20 W**.

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