Building Bridges

The questions below are due on Monday March 03, 2025; 10:00:00 PM.
 
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In this question, we'll explore a useful application of Thévenin equivalents: simplifying parts of a circuit down to make solving in the larger circuit easier. In this problem, we'll consider solving for the current labeled i in the following circuit:


 
While we could solve this using the node method directly, in this problem, we'll explore a different strategy: finding the Thévenin equivalents of two subparts of the circuit to make an equivalent circuit that's a bit easier to solve. In particular, we'll find the Thévenin equivalents of the two boxed subparts of the circuit (note that each box has a single exposed port) and then redraw the circuits with those subparts replaced with their Thévenin equivalents, and use that to solve for i.

1) Thévenin Number 1

Let's start by replacing the box on the left with its Thévenin equivalent, simplifying our circuit like so:

What are the values of v_{{\rm TH}_1} and R_{{\rm TH}_1} that would make these two circuits equivalent? Enter your answers as simple expressions, or as exact decimal numbers.

v_{{\rm TH}_1} (in Volts) =~

R_{{\rm TH}_1} (in Ohms) =~

2) Thévenin Number 2

Now let's similarly replace the box on the right with its Thévenin equivalent:

What are the values of v_{{\rm TH}_2} and R_{{\rm TH}_2} that would make these two circuits equivalent? Enter your answers as simple expressions, or as exact decimal numbers.

v_{{\rm TH}_2} (in Volts) =~

R_{{\rm TH}_2} (in Ohms) =~

3) Solving

Now, use this simplified circuit to solve for i (you should not need to do a full nodal analysis; can you simplify things further?) and enter your answer below:

i (in Amps) =~