In a circuit two or more cell of the same emf are connected in parallel in order

Options

A. Increases the pd across a resistance in the circuit
B. Decrease pd across a resistance in the circuit
C. Facilitate drawing more current from the battery system
D. Change the emf across the system of batteries.

Answer

To solve the problem of determining the effect of connecting two or more cells of the same EMF in parallel in a circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit Configuration**:
- We have a circuit with a single resistance \( R \) and a cell with EMF \( E \) and internal resistance \( r \).
- When multiple cells of the same EMF \( E \) are connected in parallel, they share the load.

2. **Analyzing the Potential Difference**:
- The potential difference across the resistance in the circuit is given by \( V = E - I \cdot r \), where \( I \) is the current flowing through the circuit.
- When cells are connected in parallel, the potential difference across each cell remains the same, which is \( E \).

3. **Evaluating the Options**:
- **Option 1**: Increase the potential difference across the resistance.
- This is incorrect because the potential difference across the resistance remains \( E \) and does not increase.
- **Option 2**: Decrease the potential difference across the resistance.
- This is also incorrect; the potential difference does not decrease.
- **Option 3**: Facilitate drawing more current from the battery system.
- This is correct. When cells are connected in parallel, the equivalent internal resistance decreases, allowing more current to flow.
- **Option 4**: Change the EMF across the system of batteries.
- This is incorrect; the EMF remains the same as that of the individual cells.

4. **Calculating Equivalent Resistance**:
- When \( n \) cells are connected in parallel, the equivalent internal resistance \( r_{eq} \) is given by:
\[ r_{eq} = \frac{r}{n} \]
- The total resistance in the circuit becomes \( R + r_{eq} \).

5. **Calculating the Current**:
- The total current \( I \) flowing from the battery can be calculated using Ohm's law:
\[ I = \frac{E}{R + r_{eq}} = \frac{E}{R + \frac{r}{n}} \]
- As \( n \) increases, \( r_{eq} \) decreases, leading to an increase in current \( I \).

### Conclusion:
The correct answer is **Option 3**: Facilitate drawing more current from the battery system.