When two resistance `10 Omega ` and `20 Omega ` are connected in series and 6 such sets are connected in parallel then the total resistance is

Options

A. `2.5 Omega `
B. `7.5 Omega `
C. `5 Omega `
D. `3.5 Omega `

Answer

To solve the problem, we need to find the total resistance when two resistances of 10 Ω and 20 Ω are connected in series, and then six such sets are connected in parallel.

### Step-by-Step Solution:

1. **Calculate the Series Resistance (Rs):**
When resistors are connected in series, the total resistance is simply the sum of the individual resistances.

\[ R_s = R_1 + R_2 \]
Here, \( R_1 = 10 \, \Omega \) and \( R_2 = 20 \, \Omega \).

\[ R_s = 10 \, \Omega + 20 \, \Omega = 30 \, \Omega \]

2. **Calculate the Parallel Resistance (Rp):**
Now, we have 6 sets of 30 Ω resistances connected in parallel. The formula for the total resistance \( R_p \) in parallel is given by:

\[ \frac{1}{R_p} = \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} \]
Since all resistances are equal, we can simplify this to:

\[ \frac{1}{R_p} = 6 \times \frac{1}{30 \, \Omega} \]

\[ \frac{1}{R_p} = \frac{6}{30} = \frac{1}{5} \]

Now, taking the reciprocal to find \( R_p \):

\[ R_p = 5 \, \Omega \]

3. **Conclusion:**
The total resistance of the entire configuration is \( R_p = 5 \, \Omega \).

### Final Answer:
The total resistance is **5 Ω**.