A cell sends a current through a resistance `R_(1)` for time `t`, next the same cell sends current through another resistance `R_(2)` for the time `t` If the same amount of heat is developed in both the resistance then find the internal resistance of the cell

Options

A. `r=sqrt(R_(1)+R_(2))`
B. `r=sqrt(R_(1)xxR_(2))`
C. `r=(R_(1)+R_(2))/(2)`
D. `r=(1)/(R_(2))+(1)/(R_(2))`

Answer

To solve the problem step by step, we will use the concepts of electric circuits, specifically focusing on the heat generated in resistors when current flows through them.

### Step 1: Understanding the Problem
We have a cell that sends a current through two different resistances \( R_1 \) and \( R_2 \) for the same amount of time \( t \). The heat generated in both resistances is the same. We need to find the internal resistance \( r \) of the cell.

### Step 2: Formula for Heat Developed
The heat developed \( H \) in a resistor when a current \( I \) flows through it for time \( t \) is given by:
\[ H = I^2 R t \]
where \( R \) is the resistance.

### Step 3: Current Through the Resistor
The current \( I \) flowing through a resistor connected to a cell with voltage \( V \) and internal resistance \( r \) can be expressed as:
\[ I = \frac{V}{R + r} \]
where \( R \) is the external resistance.

### Step 4: Heat Developed in Each Resistance
For the first resistance \( R_1 \):
\[ H_1 = I_1^2 R_1 t = \left(\frac{V}{R_1 + r}\right)^2 R_1 t \]
For the second resistance \( R_2 \):
\[ H_2 = I_2^2 R_2 t = \left(\frac{V}{R_2 + r}\right)^2 R_2 t \]

### Step 5: Setting the Heat Developed Equal
Since the heat developed in both resistances is the same:
\[ H_1 = H_2 \]
This gives us:
\[ \left(\frac{V}{R_1 + r}\right)^2 R_1 t = \left(\frac{V}{R_2 + r}\right)^2 R_2 t \]

### Step 6: Simplifying the Equation
We can cancel \( V^2 \) and \( t \) from both sides:
\[ \frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2} \]

### Step 7: Cross-Multiplying
Cross-multiplying gives:
\[ R_1 (R_2 + r)^2 = R_2 (R_1 + r)^2 \]

### Step 8: Expanding Both Sides
Expanding both sides:
\[ R_1 (R_2^2 + 2R_2 r + r^2) = R_2 (R_1^2 + 2R_1 r + r^2) \]
This simplifies to:
\[ R_1 R_2^2 + 2R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2R_2 R_1 r + R_2 r^2 \]

### Step 9: Rearranging the Equation
Rearranging gives:
\[ R_1 R_2^2 - R_2 R_1^2 + (2R_1 R_2 - 2R_2 R_1) r + (R_1 - R_2) r^2 = 0 \]
This simplifies to:
\[ R_1 R_2 (R_2 - R_1) + (R_1 - R_2) r^2 = 0 \]

### Step 10: Factoring Out Common Terms
Factoring out \( (R_1 - R_2) \):
\[ (R_1 - R_2)(r^2 + R_1 R_2) = 0 \]
Since \( R_1 \neq R_2 \), we can divide by \( (R_1 - R_2) \):
\[ r^2 = R_1 R_2 \]

### Step 11: Finding the Internal Resistance
Taking the square root gives:
\[ r = \sqrt{R_1 R_2} \]

### Final Answer
The internal resistance of the cell is:
\[ r = \sqrt{R_1 R_2} \]