A `5Omega` resistor is connected in series with a parallel combination of n resistors of `6Omega` each. The equivalent resistance is `7Omega`. Find n

Answer

To solve the problem, we need to find the value of \( n \) when a \( 5 \Omega \) resistor is connected in series with a parallel combination of \( n \) resistors of \( 6 \Omega \) each, and the equivalent resistance is \( 7 \Omega \).

### Step-by-Step Solution:

1. **Understanding the Circuit Configuration**:
- We have a \( 5 \Omega \) resistor in series with a parallel combination of \( n \) resistors, each of \( 6 \Omega \).
- The equivalent resistance of the entire circuit is given as \( 7 \Omega \).

2. **Setting Up the Equation**:
- Let the equivalent resistance of the parallel combination of \( n \) resistors be \( R_{p} \).
- The total equivalent resistance \( R_{eq} \) can be expressed as:
\[ R_{eq} = R + R_{p} = 5 + R_{p} \]
- Given that \( R_{eq} = 7 \Omega \), we can set up the equation:
\[ 7 = 5 + R_{p} \]
- From this, we can solve for \( R_{p} \):
\[ R_{p} = 7 - 5 = 2 \Omega \]

3. **Finding the Equivalent Resistance of the Parallel Resistors**:
- The formula for the equivalent resistance \( R_{p} \) of \( n \) resistors of \( 6 \Omega \) in parallel is:
\[ \frac{1}{R_{p}} = \frac{1}{6} + \frac{1}{6} + \ldots + \frac{1}{6} \quad (n \text{ times}) \]
- This can be simplified to:
\[ \frac{1}{R_{p}} = \frac{n}{6} \]
- Therefore, we can express \( R_{p} \) as:
\[ R_{p} = \frac{6}{n} \]

4. **Setting Up the Final Equation**:
- We already found that \( R_{p} = 2 \Omega \). Now we can set the two expressions for \( R_{p} \) equal to each other:
\[ \frac{6}{n} = 2 \]

5. **Solving for \( n \)**:
- To find \( n \), we can rearrange the equation:
\[ 6 = 2n \]
- Dividing both sides by \( 2 \):
\[ n = \frac{6}{2} = 3 \]

### Conclusion:
The value of \( n \) is \( 3 \).