Two electric bulbs rated `P_(1)` and `P_(2)` watt at `V` volt are connected in series across `V` volt mains then their total power consumption `P` is

Options

A. (a)`(P_(1)+P_(2))`
B. (b)`sqrt(P_(1)P_(2))`
C. (c)`P_(1)P_(2)//(P_(1)+P_(2))`
D. (d) none of these

Explanation

To solve the problem of finding the total power consumption of two electric bulbs rated \( P_1 \) and \( P_2 \) watts at \( V \) volts when connected in series, we can follow these steps:

### Step 1: Calculate the Resistance of Each Bulb
The resistance \( R \) of an electric bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
For the first bulb, the resistance \( R_1 \) is:
\[ R_1 = \frac{V^2}{P_1} \]
For the second bulb, the resistance \( R_2 \) is:
\[ R_2 = \frac{V^2}{P_2} \]

### Step 2: Calculate the Equivalent Resistance in Series
When resistors are connected in series, the total or equivalent resistance \( R_{\text{eq}} \) is the sum of the individual resistances:
\[ R_{\text{eq}} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2} \]
Factoring out \( V^2 \):
\[ R_{\text{eq}} = V^2 \left( \frac{1}{P_1} + \frac{1}{P_2} \right) \]

### Step 3: Calculate the Total Power Consumption
The total power consumption \( P \) when the bulbs are connected in series can be calculated using the formula:
\[ P = \frac{V^2}{R_{\text{eq}}} \]
Substituting the expression for \( R_{\text{eq}} \):
\[ P = \frac{V^2}{V^2 \left( \frac{1}{P_1} + \frac{1}{P_2} \right)} \]
This simplifies to:
\[ P = \frac{1}{\left( \frac{1}{P_1} + \frac{1}{P_2} \right)} \]

### Step 4: Find the Final Expression
To simplify further, we can find a common denominator:
\[ P = \frac{P_1 P_2}{P_1 + P_2} \]

### Final Answer
Thus, the total power consumption \( P \) when the two bulbs are connected in series is given by:
\[ P = \frac{P_1 P_2}{P_1 + P_2} \]

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