A group of identical cells (all in parallel) are connected to an external resistance `R_(1)`. Current through `R_(1)` is `i_(1)`. Another group of identical cells (all in series) are connected to some other resistance `R_(2)`. Current through `R_(2)` is `i_(2)`. Now one cell is removed from both the groups. Then

Options

A. `i_(1)` will decrease
B. `i_(2)` may decrease or increase depending on the value E and r
C. both (a) and (b) are correct
D. both (a) and (b) wrong

Answer

To solve the problem, we need to analyze the situation before and after removing one cell from both groups of cells connected to resistances \( R_1 \) and \( R_2 \).

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Let \( n \) be the number of identical cells in the parallel group connected to \( R_1 \).
- Let \( n' \) be the number of identical cells in the series group connected to \( R_2 \).
- The EMF of each cell is \( E \) and the internal resistance of each cell is \( r \).

2. **Calculate the Equivalent EMF and Resistance for the Parallel Group**:
- For \( n \) cells in parallel, the equivalent EMF \( E_{eq1} \) remains \( E \).
- The equivalent internal resistance \( R_{eq1} \) for \( n \) cells in parallel is given by:
\[ R_{eq1} = \frac{r}{n} \]
- The total resistance in the circuit is \( R_1 + R_{eq1} = R_1 + \frac{r}{n} \).
- The current \( i_1 \) through \( R_1 \) can be expressed using Ohm's law:
\[ i_1 = \frac{E}{R_1 + \frac{r}{n}} \]

3. **Calculate the Equivalent EMF and Resistance for the Series Group**:
- For \( n' \) cells in series, the equivalent EMF \( E_{eq2} \) is:
\[ E_{eq2} = n' E \]
- The equivalent internal resistance \( R_{eq2} \) for \( n' \) cells in series is:
\[ R_{eq2} = n' r \]
- The total resistance in the circuit is \( R_2 + R_{eq2} = R_2 + n' r \).
- The current \( i_2 \) through \( R_2 \) can be expressed as:
\[ i_2 = \frac{n' E}{R_2 + n' r} \]

4. **Removing One Cell from Each Group**:
- After removing one cell from the parallel group, the number of cells becomes \( n - 1 \):
- New equivalent internal resistance:
\[ R_{eq1}' = \frac{r}{n - 1} \]
- New current \( i_1' \):
\[ i_1' = \frac{E}{R_1 + \frac{r}{n - 1}} \]
- After removing one cell from the series group, the number of cells becomes \( n' - 1 \):
- New equivalent internal resistance:
\[ R_{eq2}' = (n' - 1) r \]
- New current \( i_2' \):
\[ i_2' = \frac{(n' - 1) E}{R_2 + (n' - 1) r} \]

5. **Analyze the Effect of Removing One Cell**:
- In both cases, removing one cell increases the equivalent internal resistance, which in turn decreases the current through the respective resistances:
- For the parallel group:
\[ i_1' < i_1 \]
- For the series group:
\[ i_2' < i_2 \]