The total resistance in the parallel combination of three resistances `9Omega, 7Omega and 5Omega`

Options

A. `1.22Omega`
B. `2.29Omega`
C. `4.22Omega`
D. `2.02Omega`

Explanation

To find the total resistance in the parallel combination of three resistances \( R_1 = 9 \, \Omega \), \( R_2 = 7 \, \Omega \), and \( R_3 = 5 \, \Omega \), we can use the formula for resistors in parallel:

\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

### Step 1: Substitute the values of the resistances into the formula.

\[ \frac{1}{R_p} = \frac{1}{9} + \frac{1}{7} + \frac{1}{5} \]

### Step 2: Calculate each term separately.

- For \( \frac{1}{9} \):
\[ \frac{1}{9} \approx 0.1111 \]

- For \( \frac{1}{7} \):
\[ \frac{1}{7} \approx 0.1429 \]

- For \( \frac{1}{5} \):
\[ \frac{1}{5} = 0.2 \]

### Step 3: Add the values together.

\[ \frac{1}{R_p} = 0.1111 + 0.1429 + 0.2 \]

Calculating the sum:
\[ \frac{1}{R_p} \approx 0.1111 + 0.1429 + 0.2 = 0.454 \]

### Step 4: Take the reciprocal to find \( R_p \).

\[ R_p = \frac{1}{0.454} \approx 2.207 \, \Omega \]

### Step 5: Round to the nearest value if necessary.

The nearest value of \( R_p \) is approximately \( 2.29 \, \Omega \).

### Final Answer:
The total resistance in the parallel combination of the three resistances is approximately \( 2.207 \, \Omega \) (or rounded to \( 2.29 \, \Omega \)).

---