Out of five resistance of `ROmega` each 3 are connected in parallel and are joined to the rest 2 in series. Find the resultant resistance.

Options

A. `(3//7)ROmega`
B. `(7//3)ROmega`
C. `(7//8)ROmega`
D. `(8//7)ROmega`

Answer

To find the resultant resistance of the given circuit, we will follow these steps:

### Step 1: Identify the Configuration
We have 5 resistors, each of resistance \( R \Omega \). Out of these, 3 resistors are connected in parallel, and the remaining 2 resistors are connected in series with the parallel combination.

### Step 2: Calculate the Equivalent Resistance of the Parallel Resistors
For resistors in parallel, the formula for equivalent resistance \( R_p \) is given by:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Since all three resistors have the same resistance \( R \):
\[ \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \]
Thus, the equivalent resistance \( R_p \) is:
\[ R_p = \frac{R}{3} \]

### Step 3: Calculate the Equivalent Resistance of the Series Resistors
The two resistors connected in series will have an equivalent resistance \( R_s \) given by:
\[ R_s = R + R = 2R \]

### Step 4: Combine the Results
Now, we combine the equivalent resistance of the parallel resistors \( R_p \) with the equivalent resistance of the series resistors \( R_s \):
\[ R_{eq} = R_p + R_s = \frac{R}{3} + 2R \]
To add these, we need a common denominator:
\[ R_{eq} = \frac{R}{3} + \frac{6R}{3} = \frac{7R}{3} \]

### Final Result
The resultant resistance of the entire circuit is:
\[ R_{eq} = \frac{7R}{3} \Omega \]