A battery of 10 cells each of e.m.f. E=1.5V and internal resistance `0.5Omega` has 1 cell wrongly connected. It is being charged by 220V power supply with an external resistance of `47Omega` in series. The potential difference across the battery.

Options

A. `2Av`
B. `2Av^2`
C. `2Av^(1/2)`
D. Can't be determined

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the total EMF of the battery
Since one cell is wrongly connected, we have 9 cells connected correctly. The EMF of each cell is given as \( E = 1.5 \, \text{V} \).

\[ \text{Total EMF} = 9 \times E = 9 \times 1.5 = 13.5 \, \text{V} \]

### Step 2: Calculate the total internal resistance of the battery
The internal resistance of each cell is given as \( r = 0.5 \, \Omega \). Since one cell is wrongly connected, we have 9 cells contributing to the internal resistance.

\[ \text{Total Internal Resistance} = 9 \times r = 9 \times 0.5 = 4.5 \, \Omega \]

### Step 3: Calculate the total resistance in the circuit
The external resistance \( R_{ext} \) is given as \( 47 \, \Omega \). The total resistance in the circuit \( R_{total} \) is the sum of the external resistance and the internal resistance of the battery.

\[ R_{total} = R_{ext} + R_{internal} = 47 + 4.5 = 51.5 \, \Omega \]

### Step 4: Calculate the total voltage supplied by the power supply
The voltage supplied by the power supply is \( V_{supply} = 220 \, \text{V} \).

### Step 5: Calculate the current flowing through the circuit
Using Ohm's law, the current \( I \) can be calculated as:

\[ I = \frac{V_{supply}}{R_{total}} = \frac{220}{51.5} \approx 4.27 \, \text{A} \]

### Step 6: Calculate the potential difference across the battery
The potential difference \( V \) across the battery can be calculated using the formula:

\[ V = \text{Total EMF} - I \times R_{internal} \]

Substituting the values we have:

\[ V = 13.5 - (4.27 \times 4.5) \approx 13.5 - 19.215 \approx -5.715 \, \text{V} \]

Since the potential difference is negative, it indicates that the battery is being charged in reverse due to the wrongly connected cell.

### Final Answer
The potential difference across the battery is approximately \( -5.715 \, \text{V} \).

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