Three equal resistor connected in series across a source of enf together dissipate `10 Watt`. If the same resistors aer connected in parallel across the same emf, then the power dissipated will be

Options

A. `10/3`
B. `10`
C. `30`
D. `90`

Answer

To solve the problem step by step, we will analyze the two cases: when the resistors are connected in series and when they are connected in parallel.

### Step 1: Determine the equivalent resistance in series
When three equal resistors \( R \) are connected in series, the equivalent resistance \( R_s \) is given by:
\[ R_s = R + R + R = 3R \]

**Hint:** Remember that in a series connection, resistances simply add up.

### Step 2: Use the power formula for series connection
The power \( P \) dissipated in a resistor is given by the formula:
\[ P = \frac{V^2}{R} \]
For the series connection, the power dissipated is:
\[ P_s = \frac{V^2}{R_s} = \frac{V^2}{3R} \]
According to the problem, this power is given as 10 Watts:
\[ \frac{V^2}{3R} = 10 \]

**Hint:** This equation relates the voltage, resistance, and power in the series configuration.

### Step 3: Rearranging to find \( V^2 \)
From the equation \( \frac{V^2}{3R} = 10 \), we can rearrange it to find \( V^2 \):
\[ V^2 = 30R \]

**Hint:** Isolate \( V^2 \) to express it in terms of \( R \).

### Step 4: Determine the equivalent resistance in parallel
When the same resistors are connected in parallel, the equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_p = \frac{R}{3} \]

**Hint:** For parallel resistors, use the reciprocal formula to find the equivalent resistance.

### Step 5: Use the power formula for parallel connection
Now, we can find the power dissipated when the resistors are connected in parallel:
\[ P_p = \frac{V^2}{R_p} = \frac{V^2}{\frac{R}{3}} = 3 \frac{V^2}{R} \]

**Hint:** The power formula remains the same, but the equivalent resistance changes.

### Step 6: Substitute \( V^2 \) into the power equation for parallel
Substituting \( V^2 = 30R \) into the power equation for the parallel configuration:
\[ P_p = 3 \cdot \frac{30R}{R} = 3 \cdot 30 = 90 \text{ Watts} \]

**Hint:** Substitute the value of \( V^2 \) from the series case into the parallel power formula.

### Final Answer
Thus, the power dissipated when the resistors are connected in parallel is:
\[ \boxed{90 \text{ Watts}} \]