To find the equivalent resistance $R_{eq}$ between points a and b in the given circuit, we need to systematically combine the resistors using series and parallel combinations.

Step-by-Step Analysis

1. Start with the right side of the circuit:

   • There are three resistors: $4 \, \Omega$, $25 \, \Omega$, and $12 \, \Omega$.
   • The $4 \, \Omega$ and $25 \, \Omega$ resistors are in series: $R_{series1} = 4 \, \Omega + 25 \, \Omega = 29 \, \Omega$
   • This $29 \, \Omega$ resistor is in parallel with the $12 \, \Omega$ resistor: $R_{parallel1} = \left( \frac{1}{29 \, \Omega} + \frac{1}{12 \, \Omega} \right)^{-1} = \frac{1}{\left( \frac{1}{29} + \frac{1}{12} \right)} \approx 8.79 \, \Omega$

2. Simplify the next combination:

   • Now, the $8.79 \, \Omega$ is in series with the $144 \, \Omega$ resistor: $R_{series2} = 8.79 \, \Omega + 144 \, \Omega = 152.79 \, \Omega$

3. Combine with the diagonal resistor:

   • The $152.79 \, \Omega$ from step 2 is in parallel with the $12 \, \Omega$ resistor (diagonal resistor): $R_{parallel2} = \left( \frac{1}{152.79 \, \Omega} + \frac{1}{12 \, \Omega} \right)^{-1} \approx 11.26 \, \Omega$

4. Next combine the left side:

   • The $11.26 \, \Omega$ is now in series with the $5.6 \, \Omega$ resistor: $R_{series3} = 11.26 \, \Omega + 5.6 \, \Omega = 16.86 \, \Omega$

5. Simplify the left branch:

   • The next combination to look at involves the $2.5 \, \Omega$ resistor, which is in series with the $16.86 \, \Omega$: $R_{series4} = 2.5 \, \Omega + 16.86 \, \Omega = 19.36 \, \Omega$

6. Work on the middle branch:

   • The $19.36 \, \Omega$ is in parallel with the $15 \, \Omega$ resistor: $R_{parallel3} = \left( \frac{1}{19.36 \, \Omega} + \frac{1}{15 \, \Omega} \right)^{-1} \approx 8.36 \, \Omega$

7. Finish the simplification:

   • Now, the $8.36 \, \Omega$ is in series with the $14 \, \Omega$ resistor: $R_{series5} = 8.36 \, \Omega + 14 \, \Omega = 22.36 \, \Omega$

8. Final combination:

   • This $22.36 \, \Omega$ resistor is in parallel with the combination of the $60 \, \Omega$ and $10 \, \Omega$ resistors in series: $R_{series6} = 60 \, \Omega + 10 \, \Omega = 70 \, \Omega$
   • The final equivalent resistance is: $R_{eq} = \left( \frac{1}{22.36 \, \Omega} + \frac{1}{70 \, \Omega} \right)^{-1} \approx 17.32 \, \Omega$

Thus, the equivalent resistance between points a and b is approximately $\boxed{17.32 \, \Omega}$.

Would you like more details on any step or a deeper explanation of the series and parallel resistor rules?

Related Questions:

1. How do you calculate the total current in a circuit if you know the equivalent resistance and the total voltage?
2. What is the difference between resistors in series and in parallel?
3. How can you use Kirchhoff's laws to solve for currents in complex circuits like this one?
4. What is the effect of adding more resistors in series on the total resistance?
5. How does combining resistors in parallel affect the total resistance?

Tip:

When combining resistors in parallel, always check if the resistances are very different. In that case, the total equivalent resistance will be closer to the smaller resistance.