Two electric bulbs A and B are disigned for the same voltage. Their power ratings are `P _(A) and P _(B)` respectively with `P _(A) lt P _(B).` If they are joined in series across a V-volt supply

Options

A. A will draw more power than B
B. B will dray more power than A
C. A and B will draw the same power
D. nothing can be decided

Answer

To solve the problem regarding the power consumption of two electric bulbs A and B connected in series across a V-volt supply, we will follow these steps:

### Step 1: Understand the relationship between power, voltage, and resistance.
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.

### Step 2: Determine the resistance of each bulb.
Since both bulbs are designed for the same voltage and we know their power ratings:
- For bulb A:
\[ P_A = \frac{V^2}{R_A} \implies R_A = \frac{V^2}{P_A} \]
- For bulb B:
\[ P_B = \frac{V^2}{R_B} \implies R_B = \frac{V^2}{P_B} \]

Given that \( P_A < P_B \), it follows that:
\[ R_A > R_B \]
This means that bulb A has a higher resistance than bulb B.

### Step 3: Analyze the series circuit.
In a series circuit, the same current \( I \) flows through both bulbs. The power consumed by each bulb can be expressed as:
- Power consumed by bulb A:
\[ P'_A = I^2 R_A \]
- Power consumed by bulb B:
\[ P'_B = I^2 R_B \]

### Step 4: Compare the power consumed by each bulb.
Since we established that \( R_A > R_B \), we can conclude:
\[ P'_A = I^2 R_A > I^2 R_B = P'_B \]
This indicates that bulb A consumes more power than bulb B.

### Conclusion:
Thus, the final conclusion is that bulb A will draw more power than bulb B when connected in series across the same voltage supply.

### Summary of the Answer:
- **Correct Statement**: A will draw more power than B.

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