Two equal resistances are connected in seires across a battery and consume a power P If these are connected in parallel, then power consumed will be

Options

A. `4.0`P
B. `2.0`
C. `P /2`
D. `P/4`

Answer

To solve the problem, we need to find the power consumed when two equal resistances are connected in parallel after knowing the power consumed when they are connected in series. Let's denote the resistance of each resistor as \( R \).

### Step-by-Step Solution:

1. **Calculate the Equivalent Resistance in Series:**
When two resistances \( R \) are connected in series, the equivalent resistance \( R_s \) is given by:
\[ R_s = R + R = 2R \]

2. **Calculate the Power in Series:**
The power consumed \( P \) in a circuit is given by the formula:
\[ P = \frac{V^2}{R_s} \]
Substituting \( R_s \):
\[ P = \frac{V^2}{2R} \]

3. **Calculate the Equivalent Resistance in Parallel:**
When the same two resistances \( R \) are connected in parallel, the equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2} \]

4. **Calculate the Power in Parallel:**
The power consumed \( P' \) in the parallel configuration is:
\[ P' = \frac{V^2}{R_p} \]
Substituting \( R_p \):
\[ P' = \frac{V^2}{\frac{R}{2}} = \frac{2V^2}{R} \]

5. **Relate \( P' \) to \( P \):**
From the previous steps, we have:
\[ P = \frac{V^2}{2R} \quad \text{and} \quad P' = \frac{2V^2}{R} \]
We can express \( P' \) in terms of \( P \):
\[ P' = 4 \left(\frac{V^2}{2R}\right) = 4P \]

### Final Answer:
Thus, when the two equal resistances are connected in parallel, the power consumed will be:
\[ \boxed{4P} \]