Tough Circuit Problem, Find Current When Switch is Closed

In summary: Therefore the equivalent resistance is easily calculated, and the answer to (d) follows.In summary, the conversation discusses a series of questions involving Kirchhoff's Loop and Junction rules, as well as equivalent resistance in series and parallel circuits. The participants discuss questions a) and b) and the difficulties they encounter in solving them. They then move on to question c) and describe their process for solving it, including the use of multiple currents and loop equations. Question d) is also discussed, with a suggestion to use the delta-Y transform. Finally, they mention the use of Thevenin equivalent circuits to solve question c) and the subsequent calculation of the equivalent resistance for question d).
  • #1
wadawalnut
14
0

Homework Statement


Qw0jA75.png

I got questions a) and b), but I'm stuck at c) and d).

Homework Equations


Kirchhoff's Loop and Junction rules.
Equivalent resistance in series: Req = R1 + R2
Equivalent resistance in parallel: Req = ((R1)-1 + (R2)-1)-1

The Attempt at a Solution


I really haven't gotten anywhere on question d), I cannot figure out how to reduce it and I can't tell which resistors are in series or parallel.

My procedure for question c) started with defining multiple currents: I had I1 going to the right through the 3 ohm resistor in the top left, I2 going to the right through the 6 ohm resistor on the left, I3 going downward through the switch (and the 3 ohm resistor underneath it). I also figured that the currents going through the 6 ohm and 3 ohm resistors on the right must be I1 and I2, because the wires that are directly connected to the battery should have the same current, and you can see that current splitting into I1 and I2 on the left. Since the resistors are symmetrical, I think on the right the current must split into I1 and I2 as well.
I then used the junction rule to find that I3 = I1 + I2, from the junction on the top in the middle.
Next, I made two loop equations, and this is where I really encounter a problem:
If you take the two symmetrical loops, like one loop being the loop on the left of the switch and the other being the loop on the right of the switch, you have the same amount of equations as unkowns, because you can write I3 in terms of I1 and I2. However, both loop equations will be written entirely in terms of unknown currents, which therefore makes a homogeneous linear system in which the only system is I1 = I2 = I3 = 0.
I realized that this occurs independent of the values of the resistors and currents, so I took the loop that goes from the battery to the bottom 6 ohm resistor, to the bottom 3 ohm resistor, and back to the battery.
The loop equation, going clockwise, is:
36 - 6I2 + 3I1 = 0
Then, taking the top left loop as described above, going clockwise, I get:
-3I1 - 3(I1 + I2) + 6I2 = 0
Which reduces to:
-6I1 + 3I2 = 0
Solving both loops simultaneously, I get:
I1 = 4A
I2 = 8A
I3 = 12A

However, the right answer is I3 = 1.71A
The answer for d) is 4.21 ohms.
 
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  • #2
Just so we're on the same page:

For a) when the switch is open, ##36.0 V## must drop across both parallel branches. You can use voltage dividers to obtain the voltage across the left hand ##3.00 \Omega## and ##6.00 \Omega## resistors. Then linking b to a with a voltage arrow and applying KVL will give ##V_{ab}## where ##a## is at a higher potential.

For b), it's simply series, series then parallel.

For c) when the switch closes, I would recommend defining two loop currents ##i_1, i_2## and all of the polarities on the resistors. Don't forget the current through the middle resistor is actually ##i_1 - i_2##. Then solve for the loop currents using mesh analysis.

For d), are you familiar with the ##\Delta - Y## transform? It would be pretty useful here.
 
  • #4
If you know about Thevenin equivalent circuits, you can do (c) in your head: 1 5/7 A. Otherwise, use Kirchoff's.
 
  • #5
After you have finished (c), you're in a position to write values for currents in all resistors. The equivalent resistance of the lot can then be calculated as 36V divided by the current drawn from that 36V source. That current comprises two currents whose values you already know (or, at least, can easily calculate).
 
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1. How do I approach solving a tough circuit problem?

When dealing with a tough circuit problem, it is important to break it down into smaller, more manageable parts. This can involve simplifying the circuit by using equivalent resistances or applying Kirchhoff's laws. It is also helpful to draw a circuit diagram and label all known values.

2. What is the significance of a switch being closed in a circuit?

A closed switch in a circuit indicates that the circuit is complete and current can flow through it. This allows for the calculation of current, voltage, and resistance in the circuit.

3. How can I find the current in a circuit when the switch is closed?

To find the current in a circuit when the switch is closed, you can use Ohm's law (I = V/R) or Kirchhoff's laws. Start by labeling all known values in the circuit and then apply the appropriate equations to solve for the current.

4. What are some common mistakes to avoid when solving a tough circuit problem?

Some common mistakes when solving a tough circuit problem include forgetting to incorporate all elements of the circuit, using incorrect values for resistors, and not properly labeling the direction of current flow. It is important to double check all calculations and make sure they align with the circuit diagram.

5. What are some tips for finding the most efficient solution to a tough circuit problem?

To find the most efficient solution to a tough circuit problem, try to simplify the circuit as much as possible using equivalent resistances and applying Kirchhoff's laws. It can also be helpful to approach the problem from different angles and to use multiple methods to confirm the accuracy of your solution.

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