A set of `4` cell each of emf `2V` and internal resistance `1.5 Omega` are connected across an external load of `10 Omega` with 2 rows, two cells in each row. Calculate the current in each row and potential difference across `10 Omega`

Answer

To solve the problem, we need to analyze the circuit configuration and calculate the current in each row of cells and the potential difference across the external load of 10Ω.

### Step 1: Understand the Circuit Configuration
We have 4 cells, each with an EMF of 2V and an internal resistance of 1.5Ω. The cells are arranged in two rows, with two cells in series in each row. The two rows are then connected in parallel across a 10Ω external load.

### Step 2: Calculate the Equivalent EMF and Internal Resistance for Each Row
Since the two cells in each row are in series, the equivalent EMF (E_eq) for each row is:
\[ E_{eq} = E_1 + E_2 = 2V + 2V = 4V \]
The internal resistance (R_eq) for two cells in series is:
\[ R_{eq} = R_1 + R_2 = 1.5Ω + 1.5Ω = 3Ω \]
Thus, each row can be represented as a single battery of 4V and 3Ω.

### Step 3: Calculate the Total Resistance in the Circuit
Now, we have two rows of batteries (each with 4V and 3Ω) connected in parallel with the external load of 10Ω. The total internal resistance of the two rows in parallel is given by:
\[ R_{total} = \frac{R_{eq1} \times R_{eq2}}{R_{eq1} + R_{eq2}} = \frac{3Ω \times 3Ω}{3Ω + 3Ω} = \frac{9Ω}{6Ω} = 1.5Ω \]

### Step 4: Calculate the Total EMF in the Circuit
Since both rows have the same EMF, the total EMF in the circuit remains:
\[ E_{total} = 4V \]

### Step 5: Calculate the Total Resistance in the Circuit
The total resistance in the circuit is the sum of the equivalent internal resistance and the external load:
\[ R_{total} = R_{internal} + R_{external} = 1.5Ω + 10Ω = 11.5Ω \]

### Step 6: Calculate the Total Current in the Circuit
Using Ohm's law, the total current (I) in the circuit can be calculated as:
\[ I = \frac{E_{total}}{R_{total}} = \frac{4V}{11.5Ω} \approx 0.3478A \]

### Step 7: Calculate the Current in Each Row
Since the two rows are in parallel, the current splits equally between them. Therefore, the current in each row (I_row) is:
\[ I_{row} = \frac{I}{2} = \frac{0.3478A}{2} \approx 0.1739A \]

### Step 8: Calculate the Potential Difference Across the External Load
The potential difference (V_load) across the external load can be calculated using Ohm's law:
\[ V_{load} = I \times R_{external} = 0.3478A \times 10Ω \approx 3.478V \]

### Final Answers
- Current in each row: \( \approx 0.1739A \)
- Potential difference across the 10Ω load: \( \approx 3.478V \)