The p.d. across the terminals of a battery is 50 V when 11 A are drawn and 60 V when 1A is drawn. The emf and internal resistance of the battery are

Options

A. A) `62 BV ,2 Omega `
B. B) `62V ,1Omega `
C. C) `61 V , 1 Omega `
D. D) `64 V , 2 Omega `

Answer

To solve the problem, we will use the information given about the potential difference (p.d.) across the terminals of the battery at two different currents. We will derive the equations based on Ohm's law and the concept of electromotive force (emf) and internal resistance.

### Step-by-Step Solution:

1. **Identify the given values**:
- When the current (I1) is 11 A, the terminal voltage (V1) is 50 V.
- When the current (I2) is 1 A, the terminal voltage (V2) is 60 V.

2. **Write the equations using the formula for terminal voltage**:
The terminal voltage (V) can be expressed as:
\[ V = E - I \cdot r \]
where \(E\) is the emf of the battery, \(I\) is the current, and \(r\) is the internal resistance.

For the first case (I1 = 11 A):
\[ 50 = E - 11r \quad \text{(Equation 1)} \]

For the second case (I2 = 1 A):
\[ 60 = E - 1r \quad \text{(Equation 2)} \]

3. **Rearrange the equations**:
From Equation 1:
\[ E = 50 + 11r \quad \text{(Equation 3)} \]

From Equation 2:
\[ E = 60 + r \quad \text{(Equation 4)} \]

4. **Set the two expressions for E equal to each other**:
\[ 50 + 11r = 60 + r \]

5. **Solve for r**:
Rearranging gives:
\[ 11r - r = 60 - 50 \]
\[ 10r = 10 \]
\[ r = 1 \, \text{ohm} \]

6. **Substitute r back to find E**:
Substitute \(r = 1\) into Equation 4:
\[ E = 60 + 1 = 61 \, \text{V} \]

7. **Final values**:
- The emf \(E\) of the battery is 61 V.
- The internal resistance \(r\) of the battery is 1 ohm.

### Conclusion:
The emf of the battery is 61 V and the internal resistance is 1 ohm.