Need Help With These Resistor Network Problems Please

[stiwi_xyz]

Homework Statement

I've been trying to figure this out for a while now so can someone give me a few pointers on this?

Relevant Equations

KVL, KCL, Ohm's Law

Consider the resistor network shown in Figure 1-1, where 𝑅1=2𝑘Ω and 𝑅2=5𝑘Ω.

(a) Calculate the numerical value of the single equivalent resistance, in 𝑘Ω, as seen from the terminals 𝐴−𝐴′. Express your answer to two decimal places.Next, consider the circuit shown in Figure 1-2, where 𝑉=30𝑉 and 𝑅3=3𝑘Ω:

This circuit can be reduced into its Thevenin equivalent, shown in Figure 1-3:

(b) Calculate the numerical value of the single Thevenin voltage, in Volts, as seen from the terminals 𝐵−𝐵′.(c) Calculate the numerical value of the single Thevenin resistance, in 𝑘Ω, as seen from the terminals 𝐵−𝐵′.

 

[etotheipi]

It is required that you show a first attempt at the question. To start you off, in part a), how much current must flow through $R_2$? Given this fact, what can you do to simplify the circuit?

 

[stiwi_xyz]

It is required that you show a first attempt at the question. To start you off, in part a), how much current must flow through R2? Given this fact, what can you do to simplify the circuit?

We can definitely simplify by using the parallel and series resistor rules but in this question specifically, I'm not sure of which ones are arranged in a parallel or series orientation.

 

[etotheipi]

We can definitely simplify by using the parallel and series resistor rules but in this question specifically, I'm not sure of which ones are arranged in a parallel or series orientation.

The top circuit is an example of what's called a balanced Wheatstone bridge. By symmetry considerations, what must be current through, or voltage across, $R_2$?

 

[stiwi_xyz]

The top circuit is an example of what's called a balanced Wheatstone bridge. By symmetry considerations, what must be current through, or voltage across, $R_2$?

I'm going to say 0, probably because its going both ways and if the resistance is equal on both sides, then the voltages should cancel out

 

[etotheipi]

I'm going to say 0, probably because its going both ways and if the resistance is equal on both sides, then the voltages should cancel out

Right! Now if zero current flows through a branch, you can remove it from the schematic. That leaves you with two parallel branches, can you calculate the effective resistance now?

 

[stiwi_xyz]

Right! Now if zero current flows through a branch, you can remove it from the schematic. That leaves you with two parallel branches, can you calculate the effective resistance now?

So I end up with an equivalent 4𝑘Ω resister on both sides and I can just do (1/4𝑘Ω) + (1/4𝑘Ω) and I get an equivalent 0.5𝑘Ω resistor?

 

[etotheipi]

So I end up with an equivalent 4𝑘Ω resister on both sides and I can just do (1/4𝑘Ω) + (1/4𝑘Ω) and I get an equivalent 0.5𝑘Ω resistor?

Remember it's $\frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{R_{eff}}$, i.e. you need to take the reciprocal. Alternatively, you can just divide the $4\text{k}\Omega$ by two since the branches are identical.

 

[stiwi_xyz]

Remember it's $\frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{R_{eff}}$, i.e. you need to take the reciprocal. Alternatively, you can just divide the $4\text{k}\Omega$ by two since the branches are identical.

All right so I got to do (1/((1/4)+(1/4)) and that gets me 2𝑘Ω for part a.

 

[etotheipi]

OK; have you come across how to calculate the Thevenin equivalent circuit before? How far have you gotten?

 

[stiwi_xyz]

OK; have you come across how to calculate the Thevenin equivalent circuit before? How far have you gotten?

I mean I know what it is, but I haven't ever run into it. I'm pretty new to electrical engineering cause I've mostly just stuck to programming until recently.

 

[etotheipi]

Essentially, the equivalent voltage is just the voltage between B and B' (in open circuit). If you can work out the current through the triangle, that should let you work this out.

To calculate the equivalent resistance, you need to replace the voltage source with a wire and calculate the resistance between B and B'.

 

[stiwi_xyz]

Essentially, the equivalent voltage is just the voltage between B and B' (in open circuit). If you can work out the current through the triangle, that should let you work this out.

To calculate the equivalent resistance, you need to replace the voltage source with a wire and calculate the resistance between B and B'.

So if the triangle is basically a loop, doesn't that mean that the current through it should add up to 0

 

[etotheipi]

So if the triangle is basically a loop, doesn't that mean that the current through it should add up to 0

It won't add up to zero, but it will be the same throughout the triangle.

 

[stiwi_xyz]

It won't add up to zero, but it will be the same throughout the triangle.

so the current would be 10amps? Using Ohm's Law I can find the current by dividing 30 by 3.

 

[etotheipi]

That's not right. If you apply Kirchoff's voltage law around the triangular loop you end up with$$V - iR_3 - iR_3 - iR_3 = 0 \implies V = 3iR_3$$So we would have $i = 30V/9k\Omega$. Also don't forget to include the "kilo" part!

 

[stiwi_xyz]

That's not right. If you apply Kirchoff's voltage law around the triangular loop you end up with$$V - iR_3 - iR_3 - iR_3 = 0 \implies V = 3iR_3$$So we would have $i = 30V/9k\Omega$. Don't forget to include the "kilo" part!

Yea so that's what I think I meant before when I said that the currents added up to 0, I just forgot that u'd have to also multiply them by the resistances. So now that we have our current of 3.333 amps, what can we do with that to actually find the overall resistance and the Thevenin voltage?

 

[etotheipi]

Yea so that's what I think I meant before when I said that the currents added up to 0, I just forgot that u'd have to also multiply them by the resistances. So now that we have our current of 3.333 amps, what can we do with that to actually find the overall resistance and the Thevenin voltage?

It should be $3.33\times 10^{-3}A$... don't forget the "kilo" part! In any case, now you just need to work out the voltage between B and B'. This is just going to be the current we just worked out multiplied by the $3k\Omega$ (N.B. That is why it is best to keep things in symbolic form, you would see the nasty numbers cancel out!)

 

[stiwi_xyz]

It should be $3.33\times 10^{-3}A$... don't forget the "kilo" part! In any case, now you just need to work out the voltage between B and B'. This is just going to be the current we just worked out multiplied by the $3k\Omega$ (N.B. That is why it is best to keep things in symbolic form, you would see the nasty numbers cancel out!)

So the Thevenin voltage is 0.003333*3, and we end up with 0.009. Now to calculate the resistance, is it possible to do it in one operation knowing the Thevenin voltage and overall current?

 

[etotheipi]

So the Thevenin voltage is 0.003333*3, and we end up with 0.009.

You should end up with 10V. Did you forget the "kilo" again!

Now to calculate the resistance, is it possible to do it in one operation knowing the Thevenin voltage and overall current?

No, you need to make sure you keep track of the quantities you have calculated. You worked out the current through the triangular section, in open circuit. That's a different thing to the current that will flow through the equivalent circuit!

You must calculate the equivalent resistance separately, replacing the voltage source by a wire.

 

[stiwi_xyz]

You should end up with 10V. Did you forget the "kilo" again!
No, you need to make sure you keep track of the quantities you have calculated. You worked out the current through the triangular section, in open circuit. That's a different thing to the current that will flow through the equivalent circuit!

You must calculate the equivalent resistance separately, replacing the voltage source by a wire.

So something similar to what we did in the first example? In this case, how would I get the overall resistance of the triangle portion?

 

[etotheipi]

So something similar to what we did in the first example? In this case, how would I get the overall resistance of the triangle portion?

It's two branches in parallel!

 

[stiwi_xyz]

It's two branches in parallel!

oh that's sneaky. So (1/((1/6)+(1/3)) and my overall resistance for the triangle is 2k Ohms now I add that to the 3k Ohms in the rest of the series and I end up with 5k Ohms?

 

[etotheipi]

That looks okay to me!

 

[stiwi_xyz]

That looks okay to me!

Yup! Everything worked out! Much thanks man.