Three resistance of values `2Omega,3Omega` and ` 6Omega` are to be connected to produce an effective resistance of `4Omega`. This can be done by connecting.

Options

A. 6Ω resistance in series with the parallel combination of 2Ω and 3Ω
B. B3Ωresistance in series with the parallel combination of 2Ω and 6Ω
C. 2Ωresistance in series with the parallel combination of 3Ω and 6Ω
D. 2Ω resistance in parallel with the parallel combination of 3Ω and 6Ω

Explanation

To find a way to connect three resistances (2Ω, 3Ω, and 6Ω) to achieve an effective resistance of 4Ω, we can follow these steps:

### Step 1: Understand the Configuration
We need to determine how to connect the resistors. The effective resistance can be achieved by connecting some resistors in parallel and others in series.

### Step 2: Connect 3Ω and 6Ω in Parallel
When resistors are connected in parallel, the formula for the equivalent resistance \( R_{eq} \) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
For our case, let’s connect the 3Ω and 6Ω resistors in parallel.

Using the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} \]

### Step 3: Calculate the Equivalent Resistance of 3Ω and 6Ω
To calculate:
\[ \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]
Thus:
\[ R_{eq} = 2Ω \]

### Step 4: Connect the 2Ω Resistor in Series
Now, we will connect the equivalent resistance (2Ω) from the parallel connection with the 2Ω resistor in series. The formula for resistors in series is:
\[ R_{total} = R_1 + R_2 \]
So, we have:
\[ R_{total} = 2Ω + 2Ω = 4Ω \]

### Conclusion
By connecting the 3Ω and 6Ω resistors in parallel and then connecting the resulting 2Ω in series with the 2Ω resistor, we achieve an effective resistance of 4Ω.

### Final Answer
The configuration to achieve an effective resistance of 4Ω is:
- Connect 3Ω and 6Ω in parallel.
- Connect the resulting equivalent resistance (2Ω) in series with the 2Ω resistor.

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