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 \( n \) given that a \( 5 \, \Omega \) resistor is in series with a parallel combination of \( n \) resistors, each of \( 6 \, \Omega \), and the equivalent resistance is \( 7 \, \Omega \).

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- We have a \( 5 \, \Omega \) resistor in series with a parallel combination of \( n \) resistors, each of \( 6 \, \Omega \).

2. **Calculate the Equivalent Resistance of the Parallel Resistors**:
- The formula for the equivalent resistance \( R_p \) of \( n \) resistors in parallel, each of resistance \( R \), is given by:
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \ldots + \frac{1}{R} \quad (n \text{ times}) \]
- For \( n \) resistors of \( 6 \, \Omega \):
\[ \frac{1}{R_p} = \frac{n}{6} \]
- Therefore, the equivalent resistance \( R_p \) is:
\[ R_p = \frac{6}{n} \]

3. **Combine the Series Resistor with the Parallel Combination**:
- The total equivalent resistance \( R_{eq} \) of the series combination of the \( 5 \, \Omega \) resistor and the parallel combination \( R_p \) is:
\[ R_{eq} = R + R_p = 5 + \frac{6}{n} \]

4. **Set Up the Equation**:
- According to the problem, the equivalent resistance is \( 7 \, \Omega \):
\[ 5 + \frac{6}{n} = 7 \]

5. **Solve for \( n \)**:
- Rearranging the equation:
\[ \frac{6}{n} = 7 - 5 \]
\[ \frac{6}{n} = 2 \]
- Cross-multiplying gives:
\[ 6 = 2n \]
- Dividing both sides by \( 2 \):
\[ n = 3 \]

### Final Answer:
Thus, the value of \( n \) is \( 3 \).