Four resistors of 6 Ω resistance each are joined as shown in the figure. The equivalent resistance of the combination is:

8 Ω

The correct answer is Option (1) → 8 Ω

Each resistor has value $6\ \Omega$.

The circuit shows that the FIRST resistor is in series with a group of THREE resistors that are all connected in parallel with each other (because both ends of these three resistors are joined by wires).

So the combination is:

• One $6\ \Omega$ resistor in series • Followed by three $6\ \Omega$ resistors in parallel

Parallel of three equal resistors:

$R_{p} = \frac{6}{3} = 2$

Now add the series resistor:

$R_{eq} = 6 + 2$

$R_{eq} = 8$

The equivalent resistance of the combination is $8\ \Omega$.