A battery of four cells in series, each having an emf of `1.14` V and an internal resistance of `2Omega` is to be used to charge a small 2 V accumulator of negligible internal resistance. What is the charging current?

Options

A. `0.1A`
B. `0.2A`
C. `0.3A`
D. `0.45A`

Answer

To find the charging current from a battery of four cells in series, we can follow these steps:

### Step 1: Determine the total EMF of the battery
Each cell has an EMF of `1.14 V`, and there are `4` cells in series. The total EMF (E) can be calculated as:
\[ E = n \times \text{EMF of each cell} = 4 \times 1.14 \, \text{V} = 4.56 \, \text{V} \]

### Step 2: Determine the total internal resistance of the battery
Each cell has an internal resistance of `2 Ω`. Since they are in series, the total internal resistance (R) is:
\[ R = n \times \text{internal resistance of each cell} = 4 \times 2 \, \Omega = 8 \, \Omega \]

### Step 3: Set up the equation using Kirchhoff's Voltage Law (KVL)
When charging the 2 V accumulator, we can apply KVL around the loop. The equation can be set up as follows:
\[ E - V - I \cdot R = 0 \]
Where:
- \( E = 4.56 \, \text{V} \) (total EMF)
- \( V = 2 \, \text{V} \) (voltage across the accumulator)
- \( R = 8 \, \Omega \) (total internal resistance)
- \( I \) is the charging current we need to find.

Rearranging the equation gives:
\[ 4.56 \, \text{V} - 2 \, \text{V} - I \cdot 8 \, \Omega = 0 \]
\[ 2.56 \, \text{V} = I \cdot 8 \, \Omega \]

### Step 4: Solve for the charging current (I)
Now, we can solve for \( I \):
\[ I = \frac{2.56 \, \text{V}}{8 \, \Omega} = 0.32 \, \text{A} \]

### Conclusion
The charging current is:
\[ I = 0.32 \, \text{A} \]