# Need Help Analyzing Circuits - Power absorbed, Capacitor energy

Thread Starter

#### Zachary Hall

Joined May 9, 2016
4
Hi I'm relatively new to the forum, but definitely have been learning a lot on here, awesome community and site. Anyways I'm doing some end of the semester tests for my electrical circuits class and am pretty lost (im a mechanical major). I've attempted the first couple questions but i'm posting all 5 questions as I'm continuing to work though it. Any help, at least in general guidance on what to do is very much appreciated!

Thanks,
Zach

#### WBahn

Joined Mar 31, 2012
26,398
Your work on the first one is correct, at least as far as you've taken it.

I can't provide much more than that until I see your best attempt to solve them. Then I can look as see where you started going wrong (if you even do).

Thread Starter

#### Zachary Hall

Joined May 9, 2016
4
Thanks for your reply, I know this may be kinda sad but I'm getting stuck right here. Not sure what to do? Do I need to find I1 and I2 and how? Could I use a KVL loop? But that doesn't really make sense to me either.

Again, thank you.

#### shteii01

Joined Feb 19, 2010
4,644
Thanks for your reply, I know this may be kinda sad but I'm getting stuck right here. Not sure what to do? Do I need to find I1 and I2 and how? Could I use a KVL loop? But that doesn't really make sense to me either.

Again, thank you.
You got the total current right. You got the total supplied power right.

When you simplified all those resistors, you sort of folded them into one big resistor. Now you need to unfold them.

Like the circuit with the 2 kOhm and 3.43 kOhm resistors. The two resistors are in series. So your $$I_{Total}$$ is passing through both resistors. You know the resistance of your 2 kOhm resistor, you know the current passing through it, use it to find the power it is absorbing.

Thread Starter

#### Zachary Hall

Joined May 9, 2016
4
Thanks sir, I determined some values from what you've suggested. Now i'm really confused on this one.... Notice I found two different values for I1. Not sure what I did wrong, but I was assuming since the (8k+4k) is parallel to the 6k then they should have the same voltage (7.58v)?

#### shteii01

Joined Feb 19, 2010
4,644
Thanks sir, I determined some values from what you've suggested. Now i'm really confused on this one.... Notice I found two different values for I1. Not sure what I did wrong, but I was assuming since the (8k+4k) is parallel to the 6k then they should have the same voltage (7.58v)?
Yes, there is a big problem.

Your first simplification is 8 kOhm and 4 kOhm resistors in series. Resistors in series add up, so 8+4=12 kOhm. All is well.

Next. You have 6 kOhm resistor in parallel with 8 kOhm resistor. That is wrong. 6 kOhm resistor is in parallel with the branch that contains 8 kOhm and 4 kOhm resistors. We already determined that resistors in series add up, so that branch contains total of 12 kOhm. Therefore 6 kOhm resistor is in parallel with 12 kOhm resistor. You need to redo this part.

Also. You need to use shortcuts so you don't waste time on math. The shortcut for two parallel resistors is: If you have X resistor in parallel with Y resistor, then equivalent resistance is $$\frac{XY}{X+Y}$$

To use your problem, 6k in parallel with 12k is going to be (6k*12k)/(6k+12k)=4 kOhm

Thread Starter

#### Zachary Hall

Joined May 9, 2016
4
shteii01 thank you! I made a pretty simple mistake, thanks for pointing it out. Also I like that resistance formula. I still didn't figure out the last question. It says to solve by Thevenin's theorem, but when I try I started to do superposition to isolate the two sources, but I got lost in that part and never was able to get Thevenin's value, nor check them with Norton's method or get max RL.

Anyways thanks a lot!
Zach

#### MrAl

Joined Jun 17, 2014
8,245
Hi,

When you get to capacitors and inductors you also may want to look into the electrical mechanical analogies such as the force-current analogy, where mathematical descriptions are similar between the mechanical system and the electrical system. A few examples are mass-capacitance, spring-inductance. Friction in the mechanical system is often taken to be 1/R where R is the resistance in the electrical system.

#### crutschow

Joined Mar 14, 2008
27,208
................... Friction in the mechanical system is often taken to be 1/R where R is the resistance in the electrical system.
I don't understand that. I would think the larger the friction, the larger the resistance, so friction would be comparable to R not 1/R.

#### shteii01

Joined Feb 19, 2010
4,644
shteii01 thank you! I made a pretty simple mistake, thanks for pointing it out. Also I like that resistance formula. I still didn't figure out the last question. It says to solve by Thevenin's theorem, but when I try I started to do superposition to isolate the two sources, but I got lost in that part and never was able to get Thevenin's value, nor check them with Norton's method or get max RL.

Anyways thanks a lot!
Zach
You have a T junction there, the 3k, 4k and 6k resistors form a T. What I would do is use Delta-T transformation to change the T into Delta. That should relocate the resistors in such a way that you can simplify them.

#### MrAl

Joined Jun 17, 2014
8,245
I don't understand that. I would think the larger the friction, the larger the resistance, so friction would be comparable to R not 1/R.
Hi there Carl,

That question and/or doubt is very understandable. That's probably because we know that resistance to something always seems to result in less of something else.

The problem is, electrical resistance has nothing to do with mechanical force, at least not directly, it's just an analogy to what does have something to do with force, namely friction. We think of resistance as an impedance to current, but it doesnt really it just drops voltage. The thing with resistance though is with a constant current when the resistance goes up the voltage (v) goes up, so we have:
v=i*R

and that is perfectly fine of course.

But when we talk mechanical we are dealing with a different system, so things might change differently because the system is analogous but not exactly the same thing for the variables in use.

So for the mechanical system we loose velocity (v) not gain it, like with voltage (v) in the electrical circuit, even though we use the same letter. We loose velocity (v) with a constant force as the friction goes up, and we gain voltage (v) as the resistance goes up, so the mechanical friction must be a different concept than the resistance in a circuit.

With a constant force (force-current analogy which is a very common one) we LOOSE velocity as the friction goes up, so we must have:
v=i/F

where i is the force this time, v is the velocity, and F is the friction. This makes sense because if we only have so much force to work with then if we increase friction the speed must go down. The units for friction in this analogy would be appropriately chosen to work out perfectly in the above equation. So we could have:
F=i/v

and so F in this analogy would be in units of force per unit velocity.

newtons/meterspersecond=newtons/meterspersecond

If we take the inverse of the right hand side:
meterspersecond/newtons

then we know if we have a friction of 1 and we want to get to a speed of 1 meter per second we need 1 newton of force.

If we instead take that newtons/meterpersecond and multiply top and bottom by seconds, we get:
newton seconds per meter
which is in the units of the damping factor used in control theory.

There are different units for friction depending on what is being done, but this analogy makes sense because it takes more force to get a higher velocity with a given friction and if friction goes up with constant force then the velocity must go down.

Last edited:
Similar threads