To calculate the Thevenin resistance $R_{Th}$ for the network given between points a and b, we'll follow these steps:

1. Turn off all independent sources:

   • The 20V voltage source will be replaced by a short circuit.
   • The 6A current source will be replaced by an open circuit.

2. Simplify the resulting circuit: After deactivating the sources, the 5Ω and 20Ω resistors will remain, along with the 3Ω resistor between points a and b. We can now calculate the equivalent resistance seen between points a and b.

Step-by-step calculation:

• The 5Ω and 20Ω resistors are in parallel, so their equivalent resistance $R_{eq1}$ is: $R_{eq1} = \frac{(5 \times 20)}{(5 + 20)} = \frac{100}{25} = 4\ \Omega$
• Now, this 4Ω equivalent resistance is in series with the 3Ω resistor between points a and b, giving the total Thevenin resistance $R_{Th}$: $R_{Th} = 4\ \Omega + 3\ \Omega = 7\ \Omega$

Conclusion:

The Thevenin resistance $R_{Th}$ between points a and b is $\mathbf{7\ \Omega}$.

Would you like more details or have any questions?

Here are some related questions:

1. How would the calculation change if there were additional resistors between points a and b?
2. What would happen if the current source were kept active while calculating $R_{Th}$?
3. How would you calculate the open-circuit voltage $V_{Th}$ in this circuit?
4. Can we apply the same method to a circuit with dependent sources? How?
5. What happens if we swap the 20Ω resistor with an inductor? How does that affect the time-domain response?

Tip: When deactivating sources, always remember to short-circuit voltage sources and open-circuit current sources.