Five equal risistors when connected in series dissipted 5 W power. If they are connected in parallel, the power dissipated will be

Options

A. 125 W
B. 96 W
C. 68 W
D. 32 W

Explanation

To solve the problem step by step, we will first analyze the scenario when the resistors are connected in series and then when they are connected in parallel.

### Step 1: Calculate the equivalent resistance in series
When resistors are connected in series, the total or equivalent resistance (R_s) is the sum of all individual resistances. Given that all five resistors are equal and denoted as R, we have:

\[ R_s = R + R + R + R + R = 5R \]

### Step 2: Use the power formula for series connection
The power dissipated in a resistor is given by the formula:

\[ P = \frac{V^2}{R} \]

In the series connection, we know the power dissipated (P_s) is 5 W. Thus, we can write:

\[ P_s = \frac{V^2}{R_s} \]

Substituting R_s from Step 1:

\[ 5 = \frac{V^2}{5R} \]

### Step 3: Rearrange to find V^2
Rearranging the equation to find V^2 gives us:

\[ V^2 = 5 \times 5R = 25R \]

### Step 4: Calculate the equivalent resistance in parallel
When the same resistors are connected in parallel, the equivalent resistance (R_p) can be calculated using the formula:

\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \]

This simplifies to:

\[ \frac{1}{R_p} = \frac{5}{R} \]

Thus, the equivalent resistance in parallel is:

\[ R_p = \frac{R}{5} \]

### Step 5: Use the power formula for parallel connection
Now, we can find the power dissipated when the resistors are connected in parallel (P_p) using the same power formula:

\[ P_p = \frac{V^2}{R_p} \]

Substituting R_p from Step 4:

\[ P_p = \frac{V^2}{\frac{R}{5}} = 5 \times \frac{V^2}{R} \]

### Step 6: Substitute V^2 from Step 3
Now, we substitute V^2 from Step 3 into the equation for P_p:

\[ P_p = 5 \times \frac{25R}{R} = 5 \times 25 = 125 \text{ W} \]

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

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