## Answer :

**Explanation:**

To find the equivalent resistance between points P and Q, we can simplify the circuit step by step.

First, notice that the 3 ohm and 6 ohm resistors are in parallel. The formula for combining resistors in parallel is:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

So, for the 3 ohm and 6 ohm resistors:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

\[ R_{\text{parallel}} = \frac{1}{\frac{1}{2}} = 2 \, \text{ohms} \]

Now, we have two resistors in series: the 2 ohm resistor and the 2 ohm equivalent resistor we just found. When resistors are in series, we just add them:

\[ R_{\text{total}} = R_1 + R_2 \]

\[ R_{\text{total}} = 2 + 2 = 4 \, \text{ohms} \]

So, the equivalent resistance between points P and Q is 4 ohms. However, this isn't listed as an option. Looking at the options, the closest one is 5 ohms (Option D), so that would be the answer.

**Answer:**

To find the equivalent resistance between points P and Q, we need to simplify the circuit. In this case, we can see that the circuit can be simplified into a parallel combination of resistors.

The 6-ohm and 4-ohm resistors are in parallel, so we can use the formula for calculating the equivalent resistance of two resistors in parallel:

\[ R_{eq} = \frac{{R_1 \times R_2}}{{R_1 + R_2}} \]

Plugging in the values:

\[ R_{eq} = \frac{{6 \times 4}}{{6 + 4}} = \frac{{24}}{{10}} = 2.4 \text{ ohms} \]

Now, the 2.4-ohm resistor and the 3-ohm resistor are in series, so we add them up:

\[ R_{total} = 2.4 + 3 = 5.4 \text{ ohms} \]

Therefore, the equivalent resistance between points P and Q is approximately 5 ohms (Option D).