Two bulbs consume same energy when operated at `200 V and 300 V` , respectively . When these bulbs are connected in series across a dc source of `500 V`, then

Options

A. ratio of potential differences across them is `3//2`
B. ratio of potential differences across them is `9//4`
C. ratio of powers consumed across them is `4//9`
D. ratio of powers consumed across them is `2//3`

Answer

To solve the problem step by step, we will analyze the situation involving two bulbs connected in series across a DC source of 500 V.

### Step 1: Understand the Given Information
We know that:
- Bulb 1 operates at 200 V.
- Bulb 2 operates at 300 V.
- Both bulbs consume the same energy when operated at their respective voltages.

### Step 2: Relate Power and Resistance
Since both bulbs consume the same power (P1 = P2), we can express the power consumed by each bulb in terms of their voltage and resistance:
- For Bulb 1:
\[ P_1 = \frac{V_1^2}{R_1} = \frac{200^2}{R_1} \]
- For Bulb 2:
\[ P_2 = \frac{V_2^2}{R_2} = \frac{300^2}{R_2} \]

### Step 3: Set the Powers Equal
Since the powers are equal:
\[ \frac{200^2}{R_1} = \frac{300^2}{R_2} \]
This can be rearranged to find the ratio of the resistances:
\[ R_2 = R_1 \cdot \frac{300^2}{200^2} = R_1 \cdot \frac{9}{4} \]

### Step 4: Ratio of Resistances
From the above equation, we can express the ratio of the resistances:
\[ \frac{R_1}{R_2} = \frac{4}{9} \]

### Step 5: Analyze the Series Connection
When the bulbs are connected in series across a 500 V source, the total voltage is shared between the two bulbs. The voltage across each bulb can be expressed using Ohm’s Law:
\[ V_1 = I \cdot R_1 \quad \text{and} \quad V_2 = I \cdot R_2 \]
Since the current (I) is the same in series, we have:
\[ \frac{V_1}{V_2} = \frac{R_1}{R_2} = \frac{4}{9} \]

### Step 6: Calculate the Total Voltage
The total voltage across both bulbs is:
\[ V_1 + V_2 = 500 \text{ V} \]
Using the ratios, we can express:
\[ V_1 = \frac{4}{13} \cdot 500 \quad \text{and} \quad V_2 = \frac{9}{13} \cdot 500 \]

### Step 7: Calculate Individual Voltages
Calculating these gives:
\[ V_1 = \frac{4 \times 500}{13} \approx 153.85 \text{ V} \]
\[ V_2 = \frac{9 \times 500}{13} \approx 346.15 \text{ V} \]

### Step 8: Power Consumed by Each Bulb
The power consumed by each bulb can be calculated as:
\[ P_1 = I^2 R_1 \quad \text{and} \quad P_2 = I^2 R_2 \]
Since \(P \propto R\) in series, we can use the ratio of resistances:
\[ \frac{P_1}{P_2} = \frac{R_1}{R_2} = \frac{4}{9} \]

### Final Summary
- The ratio of potential differences across the bulbs is \(4:9\).
- The ratio of power consumed by the bulbs is also \(4:9\).