The internal resistance of a `2.1 V` cell which gives a current `0.2 A` through a resistance of `10 Omega`

Options

A. `0.2 Omega`
B. `0.5 Omega`
C. `0.8 Omega`
D. `1.0 Omega`

Answer

To find the internal resistance of a `2.1 V` cell that gives a current of `0.2 A` through a resistance of `10 Ω`, we can follow these steps:

### Step-by-Step Solution

1. **Identify the known values:**
- EMF of the cell (E) = `2.1 V`
- Current (I) = `0.2 A`
- External resistance (R) = `10 Ω`

2. **Use Ohm's Law:**
According to Ohm's Law, the total voltage (V) in a circuit is equal to the current (I) multiplied by the total resistance (R_total). In this case, the total resistance includes both the external resistance and the internal resistance of the cell.

\[ V = I \cdot R_{total} \]

Where:
\[ R_{total} = R + r \]

Here, `r` is the internal resistance of the cell.

3. **Set up the equation:**
We can express the total voltage as:
\[ E = I \cdot (R + r) \]
Substituting the known values:
\[ 2.1 = 0.2 \cdot (10 + r) \]

4. **Solve for r:**
First, divide both sides by `0.2`:
\[ \frac{2.1}{0.2} = 10 + r \]
This simplifies to:
\[ 10.5 = 10 + r \]

Now, isolate `r`:
\[ r = 10.5 - 10 \]
\[ r = 0.5 \, \Omega \]

5. **Conclusion:**
The internal resistance of the cell is `0.5 Ω`.