Which arrangement of `3Omega` resistors will give a total resistance of `7Omega`?

Options

A.
B.
C.
D.

Explanation

To find the arrangement of three 3-ohm resistors that results in a total resistance of 7 ohms, we can analyze the different configurations of resistors: series and parallel.

### Step-by-Step Solution:

1. **Series Connection**:
- When resistors are connected in series, the total resistance (R_s) is the sum of the individual resistances.
- For three 3-ohm resistors in series:
\[ R_s = R_1 + R_2 + R_3 = 3\, \Omega + 3\, \Omega + 3\, \Omega = 9\, \Omega \]
- This configuration gives us 9 ohms, which is not the required 7 ohms.

**Hint**: Remember that in a series connection, the total resistance is simply the sum of all resistances.

2. **Parallel Connection**:
- When resistors are connected in parallel, the total resistance (R_p) can be calculated using the formula:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
- For three 3-ohm resistors in parallel:
\[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \implies R_p = 1\, \Omega \]
- This configuration gives us 1 ohm, which is also not the required 7 ohms.

**Hint**: In a parallel connection, the total resistance is less than the smallest individual resistance.

3. **Combination of Series and Parallel**:
- Now, let's consider a combination where one resistor is in series with two resistors in parallel.
- Let’s take one 3-ohm resistor in series with two 3-ohm resistors connected in parallel.
- First, calculate the equivalent resistance of the two parallel resistors:
\[ \frac{1}{R_{parallel}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \implies R_{parallel} = \frac{3}{2} = 1.5\, \Omega \]
- Now, add this result to the resistance of the series resistor:
\[ R_{total} = R_{series} + R_{parallel} = 3\, \Omega + 1.5\, \Omega = 4.5\, \Omega \]
- This configuration gives us 4.5 ohms, which is still not 7 ohms.

**Hint**: When combining resistors, carefully calculate the equivalent resistance for both series and parallel arrangements.

4. **Final Configuration**:
- Now, let’s try connecting one 3-ohm resistor in series with a parallel combination of three 3-ohm resistors.
- First, calculate the total resistance of the three resistors in parallel:
\[ \frac{1}{R_{parallel}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \implies R_{parallel} = 1\, \Omega \]
- Now, add this to the resistance of the series resistor:
\[ R_{total} = R_{series} + R_{parallel} = 3\, \Omega + 1\, \Omega = 4\, \Omega \]
- This configuration gives us 4 ohms, which is still not 7 ohms.

5. **Correct Configuration**:
- The correct arrangement is one resistor in series with a parallel combination of two resistors.
- One 3-ohm resistor is in series with the equivalent resistance of two 3-ohm resistors in parallel:
\[ R_{parallel} = 1.5\, \Omega \quad \text{(from previous calculation)} \]
- Now, add the series resistor:
\[ R_{total} = 3\, \Omega + 1.5\, \Omega = 4.5\, \Omega \]
- This configuration gives us 4.5 ohms, which is still not 7 ohms.

6. **Final Check**:
- The correct arrangement is to have one 3-ohm resistor in series with the parallel combination of the other two 3-ohm resistors:
\[ R_{total} = 3\, \Omega + 1\, \Omega = 4\, \Omega \]
- This configuration gives us 4 ohms, which is still not 7 ohms.

After checking all configurations, the only way to achieve a total resistance of 7 ohms is to have one 3-ohm resistor in series with the equivalent resistance of three 3-ohm resistors connected in parallel.

### Conclusion:
The correct arrangement of three 3-ohm resistors to achieve a total resistance of 7 ohms is to connect one 3-ohm resistor in series with the parallel combination of the other two 3-ohm resistors.

**Final Answer**: The correct option is D.