Two bulbs consume the same power when operated at 200 V and 300 V respectively. When these bulbs are connected in series across a DC source of 400 V, then the ratio of power consumed across them is

Options

A. `(2)/(3)`
B. `(3)/(2)`
C. `(4)/(9)`
D. `(9)/(4)`

Answer

To solve the problem, we need to find the ratio of power consumed by two bulbs when connected in series across a DC source of 400 V. The bulbs consume the same power when operated at 200 V and 300 V respectively.

### Step-by-Step Solution:

1. **Understanding Power Consumption**:
The power consumed by an electrical device can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( P \) is the power, \( V \) is the voltage, and \( R \) is the resistance.

2. **Setting Up the Equations**:
For the first bulb (operating at 200 V):
\[ P_1 = \frac{200^2}{R_1} \]
For the second bulb (operating at 300 V):
\[ P_2 = \frac{300^2}{R_2} \]
Since both bulbs consume the same power:
\[ \frac{200^2}{R_1} = \frac{300^2}{R_2} \]

3. **Finding the Resistance Ratio**:
Rearranging the equation gives:
\[ \frac{R_1}{R_2} = \frac{200^2}{300^2} = \left(\frac{200}{300}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
Thus, we have:
\[ R_1 : R_2 = 4 : 9 \]

4. **Connecting Bulbs in Series**:
When the bulbs are connected in series, the same current \( I \) flows through both. The total voltage across the series connection is 400 V, and the voltage across each bulb can be expressed as:
\[ V_1 = I R_1 \quad \text{and} \quad V_2 = I R_2 \]
The total voltage is:
\[ V_1 + V_2 = 400 \quad \Rightarrow \quad I R_1 + I R_2 = 400 \quad \Rightarrow \quad I(R_1 + R_2) = 400 \]

5. **Calculating Power in Series**:
The power consumed by each bulb can be expressed as:
\[ P_1 = I^2 R_1 \quad \text{and} \quad P_2 = I^2 R_2 \]
The ratio of the power consumed by the two bulbs is:
\[ \frac{P_1}{P_2} = \frac{I^2 R_1}{I^2 R_2} = \frac{R_1}{R_2} \]
Since we found \( R_1 : R_2 = 4 : 9 \), we have:
\[ \frac{P_1}{P_2} = \frac{4}{9} \]

### Final Answer:
The ratio of power consumed across the two bulbs when connected in series across a 400 V source is:
\[ \frac{P_1}{P_2} = \frac{4}{9} \]