Math Problem Statement

Calculate R_Th for the network given between a and b.

Solution

To calculate the Thevenin resistance RThR_{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 Req1R_{eq1} is: Req1=(5×20)(5+20)=10025=4 ΩR_{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 RThR_{Th}: RTh=4 Ω+3 Ω=7 ΩR_{Th} = 4\ \Omega + 3\ \Omega = 7\ \Omega

Conclusion:

The Thevenin resistance RThR_{Th} between points a and b is 7 Ω\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 RThR_{Th}?
  3. How would you calculate the open-circuit voltage VThV_{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.

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Math Problem Analysis

Mathematical Concepts

Thevenin's Theorem
Resistors in Series and Parallel
Circuit Analysis

Formulas

Parallel Resistance: R_eq = (R1 * R2) / (R1 + R2)
Series Resistance: R_total = R1 + R2

Theorems

Thevenin's Theorem

Suitable Grade Level

Undergraduate Engineering or Advanced High School