Solving Kirchoff's Rules for Triangular Array of Resistors

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In summary: You're a life saver.In summary, a triangular array of resistors is shown in the figure and a 35.0 V battery with negligible internal resistance is connected across different points (a) ab, (b) bc, and (c) ac. The key to solving this problem is to redraw the circuit as a rectangle, with the voltage source and resistor combinations in parallel and series. By using Ohm's Law and the equivalent resistance formula, the current through the circuit can be easily calculated.
  • #1
danbone87
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A triangular array of resistors is shown in the figure. What current will this array draw from a 35.0 V battery having negligible internal resistance if we connect it across (a) ab, (b) bc, (c) ac?

Kinda confused... just anything to start me off on the right track would be very helpful.
 
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  • #2
Have a good look at the circuit the triangular arrangement is probably throwing you off.
If you put a voltage source between b (positive terminal) and c (negative terminal) the c node becomes your circuit ground. So you can stretch this point and make the circuit look more rectangular. This would result in the voltage source Vbc in parallel with the 10 ohm resistor which is in series with the 15 Ohm resistor which is then in series witht he 20 Ohm resistor. Redraw the circuit that way and you should have no problem figuring out the rest of it.

hope this helps.
 
  • #3
Thanks a ton
 
  • #4
univac said:
Have a good look at the circuit the triangular arrangement is probably throwing you off.
If you put a voltage source between b (positive terminal) and c (negative terminal) the c node becomes your circuit ground. So you can stretch this point and make the circuit look more rectangular. This would result in the voltage source Vbc in parallel with the 10 ohm resistor which is in series with the 15 Ohm resistor which is then in series witht he 20 Ohm resistor. Redraw the circuit that way and you should have no problem figuring out the rest of it.

hope this helps.

This is actually a homework problem that I am also stuck on. Following the above method equals the wrong answer. I can honestly say it doesn't make sense either. That doesn't make a rectangle.
 
  • #5
johnsonc007 said:
This is actually a homework problem that I am also stuck on. Following the above method equals the wrong answer. I can honestly say it doesn't make sense either. That doesn't make a rectangle.

There are two resistors in series, which are parallel with the 3rd resistor and voltage source in each case. You should easily be able to find the equivalent resistance of the circuit and thus the current knowing this.

For a), I get 3.5A as the answer.
 
  • #6
rafehi said:
There are two resistors in series, which are parallel with the 3rd resistor and voltage source in each case. You should easily be able to find the equivalent resistance of the circuit and thus the current knowing this.

For a), I get 3.5A as the answer.

Still doesn't help. Trying to teach myself via text this due to lack of interaction from professor. Ended up buying the solution online. Would have never seen the setup, still don't follow how it becomes that.
 
  • #7
johnsonc007 said:
Still doesn't help. Trying to teach myself via text this due to lack of interaction from professor. Ended up buying the solution online. Would have never seen the setup, still don't follow how it becomes that.

Is 3.5A the correct answer?

For a), by adding a voltage source across the terminals AB, you have a voltage source in parallel with the resistor.

The other two resistors are in series with another, i.e. all current entering BC must exit through CA (and vice versa).

Because the two resistors are in series, you can add up their resistances to get the equivalent resistance (that is, you can replace the 10 and 20 ohm resistors by a single resistor of 30 ohms).

You're then left with the voltage source in parallel with the 15 ohm resistor, in parallel with the 30 ohm resistor. To find the equivalent resistance of two parallels resistors, you use:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]
which simplifies to:
[tex]R_{eq} = \frac{R_{1}*R_{2}}{R_{1}+R_{2}}[/tex]

So, Req = 15*30 / (30+15) = 10 ohms.

Then, to find the current, you simply apply Ohm's Law: i = V/R = 35/10 = 3.5A.

Does that clear it up for you? If not, which of the above steps don't you follow? It might be easier for you to visualise it as a rectangle (i.e. the AB resistor on the left, BC resistor on the top, AC resistor on the bottom, with nothing on the right line).
 
  • #8
Here's a picture showing rearrangements of the circuit for case (a). You should be able to convince yourself that the underlying topology of each version is the same -- moving components and wires around the page doesn't affect the circuit so long as all connections remain the same.
 

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  • #9
Thank you Gneil, your picture helped a ton.
 

Related to Solving Kirchoff's Rules for Triangular Array of Resistors

1. What are Kirchoff's Rules for solving a triangular array of resistors?

Kirchoff's Rules, also known as Kirchoff's Laws, are the fundamental principles used to analyze and solve electrical circuits. They are named after German physicist Gustav Kirchoff and are widely used in circuit analysis. There are two main rules: Kirchoff's Current Law (KCL) and Kirchoff's Voltage Law (KVL).

2. How do I apply Kirchoff's Current Law to a triangular array of resistors?

Kirchoff's Current Law states that the sum of all currents entering and leaving a node (or junction) in a circuit must be equal to zero. In a triangular array of resistors, you can apply this law by considering each node and writing an equation for the sum of currents entering and leaving that node. This will give you a system of equations that you can solve to find the currents in each branch of the circuit.

3. What is Kirchoff's Voltage Law and how is it used to solve a triangular array of resistors?

Kirchoff's Voltage Law states that the sum of all voltages around a closed loop in a circuit must be equal to zero. In a triangular array of resistors, you can apply this law by choosing a loop and writing an equation for the sum of voltages around that loop. This will give you a system of equations that you can solve to find the voltages across each resistor in the loop.

4. Are there any shortcuts or tricks for solving Kirchoff's Rules in a triangular array of resistors?

Yes, there are some shortcuts or techniques that can make solving a triangular array of resistors using Kirchoff's Rules easier. One technique is to start from a known value, such as the voltage or current at a specific point in the circuit, and work backwards to find the values in the rest of the circuit. Another technique is to use a matrix method to solve the system of equations, which can be especially useful for larger and more complex circuits.

5. Can Kirchoff's Rules be applied to other types of circuits besides a triangular array of resistors?

Yes, Kirchoff's Rules can be applied to any type of electrical circuit, including series circuits, parallel circuits, and more complex circuits with multiple loops and branches. The principles of KCL and KVL remain the same, and the techniques for solving the equations may vary depending on the specific circuit. However, the fundamental concepts of Kirchoff's Rules can be applied to any circuit to analyze and solve for unknown values.

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