Find the equivalent resistance at terminals a-b of each circuit in the following and apply it in Multisim program.

Steps to Solve

1. Identify the configuration of resistors

Start by analyzing the circuit and grouping the resistors into series and parallel combinations.

2. Combine the 40 Ω and 10 Ω resistors

The 40 Ω and 10 Ω resistors are in parallel. The equivalent resistance ( R_{40,10} ) can be calculated using the formula for parallel resistors:

$$ \frac{1}{R_{40,10}} = \frac{1}{40} + \frac{1}{10} $$

Calculating that gives:

$$ R_{40,10} = \frac{1}{\left(\frac{1}{40} + \frac{1}{10}\right)} = \frac{40 \times 10}{10 + 40} = \frac{400}{50} = 8 , \Omega $$

3. Combine the 60 Ω and 50 Ω resistors

Now combine the 60 Ω and 50 Ω resistors in series, as they are connected directly:

$$ R_{60,50} = 60 + 50 = 110 , \Omega $$

4. Combine the results with the 80 Ω resistor

Next, combine the equivalent resistance of the 60 Ω and 50 Ω combination (( R_{60,50} = 110 , \Omega )) in parallel with the 80 Ω resistor:

$$ \frac{1}{R_{par}} = \frac{1}{110} + \frac{1}{80} $$

Calculating gives:

$$ R_{par} = \frac{1}{\left(\frac{1}{110} + \frac{1}{80}\right)} = \frac{110 \times 80}{80 + 110} = \frac{8800}{190} \approx 46.32 , \Omega $$

5. Combine with the 30 Ω and 20 Ω resistors

Finally, add the 30 Ω resistor (in series) to the result of the parallel combination:

$$ R_{eq} = R_{par} + 30 = 46.32 + 30 = 76.32 , \Omega $$

Thus, the equivalent resistance between terminals a-b is approximately 76.32 Ω.