To calculate the equivalent resistance ($R_{\text{eq}}$) of the circuit shown:

Step-by-Step Solution:

1. Observe the Circuit: This is a network with resistors in series and parallel configurations.

2. Simplify Parallel and Series Combinations:

   • $R_2$, $R_4$, and $R_5$ are in parallel. Let’s calculate their equivalent resistance, $R_{245}$, using the formula for resistors in parallel: $\frac{1}{R_{245}} = \frac{1}{R_2} + \frac{1}{R_4} + \frac{1}{R_5}$ Since $R_2 = R_4 = R_5 = 10 \, \Omega$: $\frac{1}{R_{245}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}$ $R_{245} = \frac{10}{3} \, \Omega \, (\approx 3.33 \, \Omega)$

3. Include $R_3$ in Series:

   • The equivalent resistance of $R_{245}$ and $R_3$ in series is: $R_{345} = R_{245} + R_3 = \frac{10}{3} + 10 = \frac{40}{3} \, \Omega \, (\approx 13.33 \, \Omega)$

4. Combine $R_1$ and $R_{345}$ in Series:

   • Now, combine $R_1$ (10 $\Omega$) with $R_{345}$ in series: $R_{\text{eq}} = R_1 + R_{345} = 10 + \frac{40}{3} = \frac{70}{3} \, \Omega \, (\approx 23.33 \, \Omega)$

Final Answer:

The equivalent resistance of the circuit is approximately: $R_{\text{eq}} \approx 23.33 \, \Omega$

Do you need further clarification or a breakdown of the steps? Here are five related questions you might explore:

1. How is the total resistance calculated for different resistor combinations (series vs. parallel)?
2. What happens to the equivalent resistance if $R_2$, $R_4$, or $R_5$ are changed?
3. Can the circuit be solved using Kirchhoff’s laws instead of simplifications?
4. How does equivalent resistance change if the values of $R_1$ and $R_3$ differ significantly?
5. What tools or software can assist in solving more complex resistor networks?

Tip: Always identify resistors in series and parallel to simplify circuits step by step systematically.