The emf of a daniel cell is `1.08`V. When the terminals of the cell are connected to resistance of `3Omega`, the potential difference across the terminals is found to be 0.6 V. Then , the internal resistance of the cell is

Options

A. `1.8Omega`
B. `2.4Omega`
C. `3.24Omega`
D. `0.2Omega`

Answer

To find the internal resistance of the Daniel cell, we can follow these steps:

### Step 1: Identify the given values
- EMF of the Daniel cell (E) = 1.08 V
- Resistance connected (R) = 3 Ω
- Terminal voltage (V) = 0.6 V

### Step 2: Calculate the current flowing through the circuit
Using Ohm's law, the current (I) can be calculated using the formula:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{0.6 \, \text{V}}{3 \, \Omega} = 0.2 \, \text{A} \]

### Step 3: Use the formula relating EMF, terminal voltage, current, and internal resistance
The relationship between EMF (E), terminal voltage (V), current (I), and internal resistance (r) is given by:
\[ E = V + I \cdot r \]
Rearranging this gives:
\[ r = \frac{E - V}{I} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ r = \frac{1.08 \, \text{V} - 0.6 \, \text{V}}{0.2 \, \text{A}} \]

### Step 5: Calculate the internal resistance
Now, calculate the numerator:
\[ 1.08 - 0.6 = 0.48 \, \text{V} \]
Now, substituting this back into the equation:
\[ r = \frac{0.48 \, \text{V}}{0.2 \, \text{A}} = 2.4 \, \Omega \]

### Final Answer
The internal resistance of the cell is \( r = 2.4 \, \Omega \).

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