To reduce the circuit between terminals $a$ and $b$ to a single resistor, we need to systematically combine resistors in series and parallel until we reach an equivalent resistance for the entire circuit.

Step-by-Step Solution:

1. Identify Series and Parallel Combinations:

   • Between points $c$ and $d$, we have a combination of resistors in parallel: $2 \ \Omega$, $3 \ \Omega$, $6 \ \Omega$, and $8 \ \Omega$.
   • The $6 \ \Omega$ resistor between points $d$ and $e$ is in series with the combination found above.
   • Finally, the $16 \ \Omega$ resistor between $e$ and $b$ is also in series with the rest of the circuit.

2. Calculate the Equivalent Resistance for the Parallel Section (Between $c$ and $d$): For resistors in parallel, the equivalent resistance $R_p$ is calculated by: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}$ where $R_1 = 2 \ \Omega$, $R_2 = 3 \ \Omega$, $R_3 = 6 \ \Omega$, and $R_4 = 8 \ \Omega$.

   Plugging in the values: $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \frac{1}{8}$ Calculating each term individually: $\frac{1}{R_p} = 0.5 + 0.333 + 0.167 + 0.125 = 1.125$ Therefore: $R_p = \frac{1}{1.125} \approx 0.889 \ \Omega$

3. Combine the Equivalent Parallel Resistance with Series Resistors:

   • Now, the equivalent resistance $R_p \approx 0.889 \ \Omega$ is in series with the $6 \ \Omega$ resistor (between $d$ and $e$) and the $1 \ \Omega$ resistor (between $a$ and $c$).
   • The combined resistance from $a$ to $d$ is: $R_{acd} = 1 + 0.889 + 6 = 7.889 \ \Omega$

4. Add the Remaining Series Resistor:

   • Finally, this $7.889 \ \Omega$ is in series with the $16 \ \Omega$ resistor between points $e$ and $b$: $R_{ab} = 7.889 + 16 = 23.889 \ \Omega$

Final Answer:

The equivalent resistance between terminals $a$ and $b$ is approximately: $R_{ab} \approx 23.89 \ \Omega$

Would you like more details on any specific part of the calculation?

Here are some related questions that might help you understand this concept better:

1. How do you determine whether resistors are in series or parallel in a complex circuit?
2. What are the common techniques for simplifying resistor networks?
3. How would the equivalent resistance change if one of the resistors in the parallel combination were removed?
4. What effect does adding more parallel resistors have on the overall resistance?
5. How would you solve a similar problem if it included capacitors instead of resistors?

Tip: Always simplify parallel and series combinations gradually, one step at a time, to avoid calculation errors.