To determine the equivalent resistance of the combination of resistors $R_3$, $R_4$, $R_5$, and $R_6$, we need to analyze how these resistors are connected.

From the image:

1. Parallel Combination:
   • $R_4$ and $R_5$ are connected in series, which we can combine first: $R_{45} = R_4 + R_5$
   • $R_{45}$ is then connected in parallel with $R_6$: $R_{456} = \left(\frac{1}{R_{45}} + \frac{1}{R_6}\right)^{-1} = \left(\frac{1}{R_4 + R_5} + \frac{1}{R_6}\right)^{-1}$
   • Now, $R_3$ is connected in series with $R_{456}$: $R_{3456} = R_3 + R_{456}$
   • Thus, the equivalent resistance of the combination of $R_3$, $R_4$, $R_5$, and $R_6$ is given by: $R_{\text{eq}} = R_3 + \left(\frac{1}{R_4 + R_5} + \frac{1}{R_6}\right)^{-1}$

Would you like more details on any specific part of this process?

Here are five related questions to deepen your understanding:

1. How does the series connection of resistors differ from a parallel connection in terms of equivalent resistance?
2. If one of the resistors $R_4$ or $R_5$ were removed, how would that affect the equivalent resistance?
3. Can you calculate the equivalent resistance if specific values for $R_3$, $R_4$, $R_5$, and $R_6$ were given?
4. What are the implications of high versus low resistance in a circuit?
5. How does the equivalent resistance of a circuit impact the total current flowing through it?

Tip: Always remember to simplify resistor networks step by step, combining series and parallel connections progressively to find the overall equivalent resistance.