Five equal resistors when connected in series dissipated 5 W power. If they are connected in paralle, the power dissipated will be

Options

A. 25 W
B. 50 W
C. 100 W
D. 125 W

Explanation

To solve the problem, we need to find the power dissipated when five equal resistors are connected in parallel, given that they dissipate 5 W when connected in series.

### Step-by-Step Solution:

1. **Understanding the Series Connection**:
When resistors are connected in series, the total resistance \( R_s \) is the sum of the individual resistances. If each resistor has resistance \( R \), then:
\[ R_s = R + R + R + R + R = 5R \]

2. **Power Dissipated in Series**:
The power \( P \) dissipated in a resistor is given by the formula:
\[ P = \frac{V^2}{R} \]
For the series connection, we have:
\[ P_s = \frac{V^2}{R_s} = \frac{V^2}{5R} \]
According to the problem, \( P_s = 5 \, \text{W} \). Thus:
\[ 5 = \frac{V^2}{5R} \]
Rearranging gives:
\[ V^2 = 25R \]

3. **Understanding the Parallel Connection**:
When the same resistors are connected in parallel, the equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R} \]
Therefore:
\[ R_p = \frac{R}{5} \]

4. **Power Dissipated in Parallel**:
The power dissipated when connected in parallel is given by:
\[ P_p = \frac{V^2}{R_p} \]
Substituting \( R_p \):
\[ P_p = \frac{V^2}{\frac{R}{5}} = 5 \cdot \frac{V^2}{R} \]

5. **Substituting \( V^2 \)**:
From step 2, we know \( V^2 = 25R \). Substituting this into the power equation for parallel connection:
\[ P_p = 5 \cdot \frac{25R}{R} = 5 \cdot 25 = 125 \, \text{W} \]

### Final Answer:
The power dissipated when the resistors are connected in parallel is **125 W**.