The equivalent resistance of a group of resistance is `R`. If another resistanceis connected in parallel to the group, its new equivalent becomes `R_(1)` and if it is connected in series to the group, it new equivalent bemomes `R_(2)` we have

Options

A. `R_(1) gt R` or `R_(2) gt R`
B. `R_(1) lt R` or `R_(2) gt R`
C. `R_(1) gt R` or `R_(2) lt R`
D. `R_(1) lt R_(2)` or `R_(2) lt R`

Explanation

To solve the problem, we need to analyze the equivalent resistances when a resistance \( R \) is connected in parallel and in series with another resistance \( r \).

### Step-by-Step Solution:

1. **Understanding the Parallel Connection**:
When a resistance \( r \) is connected in parallel with \( R \), the formula for the equivalent resistance \( R_1 \) is given by:
\[ \frac{1}{R_1} = \frac{1}{R} + \frac{1}{r} \]
Rearranging this, we can express \( R_1 \):
\[ R_1 = \frac{R \cdot r}{R + r} \]
This shows that \( R_1 \) is always less than both \( R \) and \( r \).

2. **Understanding the Series Connection**:
When the same resistance \( r \) is connected in series with \( R \), the equivalent resistance \( R_2 \) is given by:
\[ R_2 = R + r \]
This indicates that \( R_2 \) is always greater than either \( R \) or \( r \).

3. **Establishing Relationships**:
From the above two equations, we can establish the following relationships:
- Since \( R_1 < R \) (from the parallel connection),
- And \( R_2 > R \) (from the series connection),
- We can conclude:
\[ R_1 < R < R_2 \]

### Final Relationships:
Thus, we have:
\[ R_1 < R < R_2 \]