Four equal resistance dissipated 5 W of power together when connected in series to a battery of negligible internal resistance . The total power dissipated in these resistance when connected in parallel across the same battery would be .

Options

A. 125W
B. 80W
C. 20W
D. 5W

Answer

To solve the problem, we need to determine the total power dissipated when four equal resistances are connected in parallel, given that they dissipate 5 W when connected in series.

### Step-by-Step Solution:

1. **Understanding the Series Configuration**:
- Let the resistance of each resistor be \( R \).
- When connected in series, the total resistance \( R_s \) is:
\[ R_s = R + R + R + R = 4R \]

2. **Using the Power Formula in Series**:
- The power dissipated in the series configuration is given as 5 W. The power \( P \) can be expressed as:
\[ P = \frac{V^2}{R_s} \]
- Substituting for \( R_s \):
\[ 5 = \frac{V^2}{4R} \]
- Rearranging gives us:
\[ V^2 = 20R \quad \text{(Equation 1)} \]

3. **Understanding the Parallel Configuration**:
- When the resistors are connected in parallel, the total resistance \( R_p \) is:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{4}{R} \implies R_p = \frac{R}{4} \]

4. **Using the Power Formula in Parallel**:
- The power dissipated in the parallel configuration is given by:
\[ P' = \frac{V^2}{R_p} \]
- Substituting for \( R_p \):
\[ P' = \frac{V^2}{\frac{R}{4}} = 4 \frac{V^2}{R} \]

5. **Substituting \( V^2 \) from Equation 1**:
- We already found \( V^2 = 20R \) from Equation 1. Substituting this into the power equation for the parallel configuration:
\[ P' = 4 \cdot \frac{20R}{R} = 4 \cdot 20 = 80 \text{ W} \]

### Final Answer:
The total power dissipated in these resistances when connected in parallel across the same battery is **80 W**.

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