To find the equivalent resistance between terminals $A$ and $B$ for the given network, we need to follow systematic steps involving combination of resistors, both in series and in parallel. The key is identifying how resistors are connected across various points.

Steps:

1. Simplify Parallel and Series Combinations:

   • Start by identifying resistors that are clearly in series or parallel and reduce them step by step.

2. Look at Series and Parallel Combinations in Different Sections:

   • The two resistors labeled $3 \, \Omega$ between $G$ and $F$ and between $F$ and $D$ are in series, so their combined resistance is: $R_{GF} = 3 \, \Omega + 3 \, \Omega = 6 \, \Omega.$

   • Now, the $6 \, \Omega$ (from $GF$) is in parallel with the $1 \, \Omega$ resistor between $F$ and $E$. The combined resistance of two resistors in parallel is given by: $\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}.$ Substituting values for the $6 \, \Omega$ and $1 \, \Omega$ resistors: $\frac{1}{R_{\text{parallel}}} = \frac{1}{6} + \frac{1}{1} = \frac{1 + 6}{6} = \frac{7}{6},$ so $R_{\text{parallel}} = \frac{6}{7} \, \Omega \approx 0.857 \, \Omega.$

3. Combine with Series Resistor:

   • Now, this $0.857 \, \Omega$ is in series with the $3 \, \Omega$ resistor between $G$ and $C$, so the total resistance between $C$ and $D$ becomes: $R_{CD} = 3 \, \Omega + 0.857 \, \Omega \approx 3.857 \, \Omega.$

4. Parallel Between $C$ and $D$ and the Rest of the Network:

   • Now, look at the two resistors connected between $C$ and $D$: one is $3.857 \, \Omega$ and the other is $1 \, \Omega$. These two resistors are in parallel, so their combined resistance is: $\frac{1}{R_{\text{total}}} = \frac{1}{3.857} + \frac{1}{1}.$ Calculating the parallel combination: $\frac{1}{R_{\text{total}}} = \frac{1}{3.857} + 1 \approx 1.259,$ so $R_{\text{total}} \approx 0.794 \, \Omega.$

5. Combine with Rest of the Circuit:

   • The $R_{\text{total}} \approx 0.794 \, \Omega$ is in series with the $8 \, \Omega$ resistor between $A$ and $C$, giving the total resistance between $A$ and $D$: $R_{AD} = 8 \, \Omega + 0.794 \, \Omega \approx 8.794 \, \Omega.$

6. Final Parallel with $7.6 \, \Omega$ Resistor:

   • Lastly, the combined $R_{AD} \approx 8.794 \, \Omega$ is in parallel with the $7.6 \, \Omega$ resistor between $A$ and $B$. Applying the parallel resistor formula: $\frac{1}{R_{\text{eq}}} = \frac{1}{8.794} + \frac{1}{7.6},$ we calculate: $\frac{1}{R_{\text{eq}}} = 0.1137 + 0.1316 \approx 0.2453,$ so $R_{\text{eq}} = \frac{1}{0.2453} \approx 4.08 \, \Omega.$

Final Answer:

The equivalent resistance between terminals $A$ and $B$ is approximately $\boxed{4.08 \, \Omega}$.

Would you like a more detailed breakdown of any specific step?

5 Related Questions:

1. What is the difference between resistors in series and parallel?
2. How does one calculate the total resistance of multiple resistors in parallel?
3. What happens to the overall resistance if resistors of unequal values are connected in parallel?
4. How can the Wheatstone Bridge principle simplify circuits like this one?
5. How would adding another resistor between points $C$ and $D$ affect the overall resistance?

Tip:

Always look for resistors in series or parallel first to simplify complex circuits step by step!