Three equal resistances connected is series across a source of e.m.f consume `20` watt. If the same resistor are connected in parallel across the same source of e.m.f., what would be the power dissipated ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We have three equal resistances (let's denote each resistance as R) connected in series across a source of EMF (E). The power consumed in this configuration is given as 20 watts.

### Step 2: Calculate the equivalent resistance in series
When resistances are connected in series, the total resistance \( R_s \) is the sum of the individual resistances:
\[ R_s = R + R + R = 3R \]

### Step 3: Use the power formula for the series connection
The power consumed in a circuit can be calculated using the formula:
\[ P = \frac{E^2}{R} \]
For the series connection, we can substitute \( R \) with \( R_s \):
\[ 20 = \frac{E^2}{3R} \]
From this, we can express \( E^2 \):
\[ E^2 = 20 \times 3R = 60R \quad \text{(Equation 1)} \]

### Step 4: Calculate the equivalent resistance in parallel
Now, we will consider the case when the same resistors are connected in parallel. The equivalent resistance \( R_p \) for three equal resistances in parallel is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \]
Thus, we can find \( R_p \):
\[ R_p = \frac{R}{3} \]

### Step 5: Use the power formula for the parallel connection
Now we will calculate the power consumed when the resistors are connected in parallel:
\[ P' = \frac{E^2}{R_p} \]
Substituting \( R_p \):
\[ P' = \frac{E^2}{\frac{R}{3}} = 3 \cdot \frac{E^2}{R} \quad \text{(Equation 2)} \]

### Step 6: Substitute \( E^2 \) from Equation 1 into Equation 2
Now we will substitute \( E^2 \) from Equation 1 into Equation 2:
\[ P' = 3 \cdot \frac{60R}{R} = 3 \cdot 60 = 180 \text{ watts} \]

### Final Answer
The power dissipated when the resistors are connected in parallel is **180 watts**.

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