A cell supplies a current of 2A to an exertnal resistance of `2Omega ` and a current of 1 ampere to another external resistance of `4.5 Omega ` connected across the cell separately . Select the correct alternative (s).

Options

A. Emf of the cell is 5 V
B. Emf of the cell is 4.25 V
C. Internal resistance of the cell is 0.5 `Omega`
D. Internal resistace of the cell is `3.25 Omega `

Answer

To solve the problem, we need to find the electromotive force (emf) \( E \) and the internal resistance \( r \) of the cell based on the given currents and external resistances.

### Step 1: Analyze the first case
In the first case, the cell supplies a current \( I_1 = 2 \, \text{A} \) to an external resistance \( R_1 = 2 \, \Omega \).

Using Ohm's law, we can express the emf \( E \) as:
\[ E = I_1 \cdot (R_1 + r) \]
Substituting the known values:
\[ E = 2 \cdot (2 + r) \]
This simplifies to:
\[ E = 4 + 2r \quad \text{(Equation 1)} \]

### Step 2: Analyze the second case
In the second case, the cell supplies a current \( I_2 = 1 \, \text{A} \) to an external resistance \( R_2 = 4.5 \, \Omega \).

Again, using Ohm's law, we can express the emf \( E \) as:
\[ E = I_2 \cdot (R_2 + r) \]
Substituting the known values:
\[ E = 1 \cdot (4.5 + r) \]
This simplifies to:
\[ E = 4.5 + r \quad \text{(Equation 2)} \]

### Step 3: Set the equations equal
Since both expressions represent the same emf \( E \), we can set Equation 1 equal to Equation 2:
\[ 4 + 2r = 4.5 + r \]

### Step 4: Solve for internal resistance \( r \)
Rearranging the equation gives:
\[ 2r - r = 4.5 - 4 \]
\[ r = 0.5 \, \Omega \]

### Step 5: Substitute \( r \) back to find \( E \)
Now that we have \( r \), we can substitute it back into either Equation 1 or Equation 2 to find \( E \). Using Equation 1:
\[ E = 4 + 2(0.5) = 4 + 1 = 5 \, \text{V} \]

### Conclusion
The emf of the cell is \( 5 \, \text{V} \) and the internal resistance is \( 0.5 \, \Omega \).

### Final Answers
- Emf \( E = 5 \, \text{V} \)
- Internal Resistance \( r = 0.5 \, \Omega \)