Utility of Equivalent Circuits

The questions below are due on Monday September 30, 2024; 10:00:00 PM.
 
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One of the practical ways that we use Thévenin and Norton equivalents is to simplify complex circuits to make them easier to solve. In this problem, we'll work through a few examples of this idea together.

1) Circuit 1

Consider solving for the variable labeled i in the circuit below:

We could certainly solve directly using the node method, but let's consider another method for now. In particular, let's try replacing everything but the 3\Omega resistor with a Thévenin equivalent:

Solving the circuit on the right is certainly easier than solving the circuit on the left! So let's go ahead and try to find the Thévenin equivalent of the red portion. In last week's p-set, we saw several different approaches we could take. Here, we'll walk through the approach of finding the open-circuit voltage (Theévenin voltage) first and then finding the Thévenin resistance directly by zeroing out the sources.

1.1) Thévenin Voltage

To find the Thévenin voltage, we leave the two terminals disconnected and measure the voltage drop across that port, which is equivalent to finding e_+ - e_- in the circuit below:

What is V_{\rm TH}, in Volts? Enter your result as a single exact number.
V_{\rm TH} (in Volts) =~

1.2) Thévenin Resistance

Now we can find the Thévenin resistance by zeroing out the source and finding the resistance between the two terminals.

Which of the following represents the network of interest with the source value zeroed out?

Given this, what is the equivalent resistance (in Ohms) between the two terminals? Enter your answer as a simple expression, or as an exact decimal number.

R_{\rm TH} =~

1.3) Solving

Now, given the Thévenin equivalent voltage and resistance you found above, solve for the current labeled i in the original circuit? Do not use the node method directly; instead, replace everything but the 3\Omega resistor with its Thévenin equivalent first.

What is the value of i, in Amps? Enter your answer as a simple expression or single exact decimal number.

i = ~$

2) Circuit 2

Now let's look at another example of a way we can simplify a circuit, by repeatedly Thévenizing/Nortonizing parts of the circuit to collapse things down. This isn't always possible, but it can be nice when it works.

Let's look at the following circuit:

We could certainly solve for v_o using superposition, but let's try a different technique here. In particular, let's try replacing just the left-most part of the circuit with its Norton equivalent. The original is shown on the left with two components highlighted, and we'd like to replace it with the components shown in red on the right:

What are I_1 (Amps) and R_1 (Ohms) in order for the two networks colored in red to be equivalent? Enter your answer as a comma-separated list of I_1, R_1, where each is a simple expression or exact number.

I_1 (Amps), R_1 (Ohms) =~

From there, we can simplify things using parallel combinations (there are two pairs of components we can simplify in this way) to arrive at a circuit like the following:

What are I_2 (Amps) and R_2 (Ohms) in order for this to be equivalent to the networks above? Enter your answer as a comma-separated list of I_2, R_2, where each is a simple expression or exact number.

I_2 (Amps), R_2 (Ohms) =~

We're almost there, but it let's take one more step, converting the red portion back into a Thévenin:

What are V_3 (Volts) and R_3 (Ohms) in order for this to be equivalent to the networks above? Enter your answer as a comma-separated list of V_3, R_3, where each is a simple expression or exact number.

V_3 (Volts), R_3 (Ohms) =~

With that last simplification done, we are left with a voltage divider, which we can use to solve for v_o. Enter your answer as a simple expression or an exact deminal number below:

v_o (in Volts) = ~