A 50 V battery is supplying current of 10 A when connected to a resistor. If the efficiency of the battery at this current is 25%, then the internal resistance of the battery is

Options

A. `2.5 Omega`
B. ` 3.75 Omega`
C. `1.25 Omega`
D. `5 Omega`

Explanation

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

### Step 1: Understand the given information
We have a battery with:
- EMF (Electromotive Force) \( V = 50 \, \text{V} \)
- Current \( I = 10 \, \text{A} \)
- Efficiency \( \eta = 25\% \)

### Step 2: Write the formula for efficiency
Efficiency (\( \eta \)) of the battery can be expressed as:
\[ \eta = \frac{\text{Power delivered to the external resistor}}{\text{Power supplied by the battery}} \times 100 \]
This can be rewritten as:
\[ \eta = \frac{P_{\text{ideal}}}{P_{\text{real}}} \]

### Step 3: Calculate power in ideal and real conditions
- The power delivered to the external resistor (ideal power) is given by:
\[ P_{\text{ideal}} = I^2 R \]
- The total power supplied by the battery (real power) is given by:
\[ P_{\text{real}} = I^2 (R + r) \]
where \( R \) is the external resistance and \( r \) is the internal resistance.

### Step 4: Set up the efficiency equation
Substituting the expressions for power into the efficiency equation, we have:
\[ \frac{I^2 R}{I^2 (R + r)} = \frac{25}{100} \]
This simplifies to:
\[ \frac{R}{R + r} = \frac{1}{4} \]

### Step 5: Cross-multiply and rearrange
Cross-multiplying gives:
\[ 4R = R + r \]
Rearranging this, we find:
\[ r = 4R - R = 3R \]

### Step 6: Use Ohm's Law to find the relationship between voltage, current, and resistance
From Ohm's Law, we know:
\[ V = I(R + r) \]
Substituting the values we have:
\[ 50 = 10(R + r) \]
This simplifies to:
\[ R + r = 5 \quad \text{(1)} \]

### Step 7: Substitute \( r = 3R \) into equation (1)
Now substituting \( r = 3R \) into equation (1):
\[ R + 3R = 5 \]
This simplifies to:
\[ 4R = 5 \]

### Step 8: Solve for \( R \)
Dividing both sides by 4 gives:
\[ R = \frac{5}{4} \, \text{ohm} \]

### Step 9: Calculate internal resistance \( r \)
Now substituting \( R \) back into the equation for internal resistance:
\[ r = 3R = 3 \times \frac{5}{4} = \frac{15}{4} \, \text{ohm} \]

### Step 10: Final answer
Thus, the internal resistance of the battery is:
\[ r = 3.75 \, \text{ohm} \]