A battery of EMF E produces currents 4 A and 3 A when connected to external resistance ` 1 Omega` and `2 Omega` respectively. The internal resistance of the battery is

Options

A. `0.5 Omega`
B. `2 Omega`
C. `1.5 Omega`
D. `1 Omega`

Answer

To find the internal resistance of the battery given the currents produced by different external resistances, we can follow these steps:

### Step 1: Understand the Problem
We have a battery with an EMF \( E \) that produces currents of 4 A and 3 A when connected to external resistances of 1 Ω and 2 Ω, respectively. We need to find the internal resistance \( r \) of the battery.

### Step 2: Set Up the Equations
Using Ohm's law, we can express the EMF in terms of the current and resistances.

1. When connected to 1 Ω:
\[ E = I_1 \cdot (R + r) = 4 \cdot (1 + r) \]
This gives us our first equation:
\[ E = 4(1 + r) \quad \text{(Equation 1)} \]

2. When connected to 2 Ω:
\[ E = I_2 \cdot (R + r) = 3 \cdot (2 + r) \]
This gives us our second equation:
\[ E = 3(2 + r) \quad \text{(Equation 2)} \]

### Step 3: Equate the Two Expressions for EMF
Since both equations represent the same EMF \( E \), we can set them equal to each other:
\[ 4(1 + r) = 3(2 + r) \]

### Step 4: Solve for \( r \)
Expanding both sides:
\[ 4 + 4r = 6 + 3r \]
Now, rearranging the equation to isolate \( r \):
\[ 4r - 3r = 6 - 4 \]
\[ r = 2 \, \Omega \]

### Conclusion
The internal resistance of the battery is \( 2 \, \Omega \).

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