Current P through a certain non-metallic rod is given as `i_(0)=0.2V^((3)/(2))` where V is potential difference across it. The rod is connected in series with a resistance to a 12 volt ideal bettery. What should be resistance ( in `Omega`) so that power dissipated in the resistance is twice that of rod.

Explanation

To solve the problem, we need to find the resistance \( R \) such that the power dissipated in the resistance is twice that of the power dissipated in the non-metallic rod. Let's go through the steps systematically.

### Step 1: Understand the given information
We are given that the current \( I_0 \) through the rod is expressed as:
\[ I_0 = 0.2 V^{3/2} \]
where \( V \) is the potential difference across the rod. The rod is connected in series with a resistance \( R \) to a 12-volt ideal battery.

### Step 2: Set up the equations for power
The total voltage from the battery is 12 volts. The voltage across the rod is \( V \), and the voltage across the resistance \( R \) is \( 12 - V \).

The power dissipated in the rod \( P_{rod} \) can be expressed as:
\[ P_{rod} = V \cdot I_0 \]
The power dissipated in the resistance \( P_R \) is given by:
\[ P_R = (12 - V) \cdot I_0 \]

### Step 3: Set up the relationship between the powers
According to the problem, the power dissipated in the resistance is twice that of the rod:
\[ P_R = 2 P_{rod} \]
Substituting the expressions for power:
\[ (12 - V) I_0 = 2 V I_0 \]

### Step 4: Simplify the equation
Since \( I_0 \) is common in both terms, we can cancel it out (assuming \( I_0 \neq 0 \)):
\[ 12 - V = 2V \]
Rearranging gives:
\[ 12 = 3V \]
Thus,
\[ V = \frac{12}{3} = 4 \text{ volts} \]

### Step 5: Calculate the current \( I_0 \)
Now, substituting \( V = 4 \) volts back into the expression for \( I_0 \):
\[ I_0 = 0.2 \cdot (4)^{3/2} \]
Calculating \( (4)^{3/2} \):
\[ (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8 \]
Thus,
\[ I_0 = 0.2 \cdot 8 = 1.6 \text{ A} \]

### Step 6: Find the resistance \( R \)
Now we can find the resistance \( R \) using the power relationship:
\[ P_R = (12 - V) I_0 = (12 - 4) \cdot 1.6 = 8 \cdot 1.6 = 12.8 \text{ W} \]
The power dissipated in the resistance can also be expressed as:
\[ P_R = I_0^2 R \]
Setting the two expressions for power equal gives:
\[ I_0^2 R = 12.8 \]
Substituting \( I_0 = 1.6 \):
\[ (1.6)^2 R = 12.8 \]
Calculating \( (1.6)^2 \):
\[ (1.6)^2 = 2.56 \]
Thus,
\[ 2.56 R = 12.8 \]
Solving for \( R \):
\[ R = \frac{12.8}{2.56} = 5 \, \Omega \]

### Final Answer
The required resistance \( R \) is:
\[ \boxed{5 \, \Omega} \]