# Homework Help: Simple Curcuitry Help

1. Feb 2, 2010

### LP20

1. The problem statement, all variables and given/known data
Given the circuit shown find:
a) value of i_a
b) value of i_b
c) value of V_o
d) power dissipated in each resistor
e) power delivered by the 50v source

Picture: (http://img197.imageshack.us/img197/4515/1000941y.jpg [Broken])

I am very new to circuits, any help would be much appreciated.

3. The attempt at a solution
I understand how to do kcl and kvl and I understand ohm's laws, but I just can't seem to find out where to start. I've tried many things. And now I'm thinking it's how I'm assigning my passive sign convention. Do I even need to assign signs?! help please!

Last edited by a moderator: May 4, 2017
2. Feb 2, 2010

### cepheid

Staff Emeritus
Hi LP20,

I think that KCL is the best way to solve this problem. Consider the "node" given by the top black dot. KCL is basically a statement of local conservation of charge (i.e. it says that electric charge can't just disappear!). What this means is that the total current entering the node must be equal to the current leaving the node. Let us call this node point '1'. Therefore, the potential at that node is v1. KCL says

current into node = current out of node

I1 = Ia + Ib

The current entering node 1, which we've labelled I1, is just the current across the branch with the 4 ohm resistor in it. The current through that resistor is just given by the potential difference (or 'voltage') across that resistor, divided by the resistance (by Ohm's law):

I1 = (50 V - V1)/(4 ohms)

The currents through the other two branches (Ia & Ib) can be expressed in terms of voltages and resistances in the circuit in a similar way. Therefore, KCL will give you a system of equations to solve, and you should be able to figure out the unknowns.

3. Feb 2, 2010

### LP20

ok thanks! maybe it's the circuit theory I'm still having trouble with. If I try to make an equation for Ia : (i=v/r) which voltage would I use? would it be the one from the source or the one from the 4 ohm branch? or would it be a totally different voltage?
i=v?/20

4. Feb 2, 2010

### cepheid

Staff Emeritus
Umm, well look at the diagram. What branch does Ia flow across? It flows across the branch with the 20 ohm resistor in it, right?

By ohm's law, the current (Ia) through the 20 ohm resistor will just depend upon the voltage across it. That's what's driving the current to go across that resistor.

What is the voltage across the 20 ohm resistor?

5. Feb 2, 2010

### LP20

Would the voltage be 50v? that's the only voltage source but doesn't voltage change once passed through a resistor? or would the voltage be unknown making this equation v=20ia?
Maybe I'm over thinking things. I feel half retarded lol. I know there is just something I'm not understanding about this.

6. Feb 2, 2010

### cepheid

Staff Emeritus
By definition, the voltage across the resistor is just the difference between the potentials at either end of it. The potential at the "top" end is just the potential at node 1, which we've labelled V1. We don't know what it is right now.

The potential at the "bottom" end is...? Well, the bottom end is connected to the minus terminal of the source. Now, recall that it really doesn't matter where we choose our reference point (from which all potentials are measured) to be. However, in situations like this, we often, for convenience, reference everything to the minus terminal of the source. In other words, this is the point that we are choosing to assign a potential of 0 volts. We call this reference point from which all potentials are measured, 'ground.'

So, one end of the resistor is at V1 and the other end is at ground (or 'grounded'). Therefore, the equation for Ia becomes:

Ia = (V1 - 0 V)/(20 ohms)

7. Feb 2, 2010

### LP20

oh ok. I believe I understand what your saying. Except I'm looking though all my notes and my book like crazy to find the part about 0v for ground and it doesn't look like we have learned it yet. No worries though, with this knowledge I will try to solve or get more equations now then solve. I will write back if another problem occurs.

8. Feb 2, 2010

### LP20

hey wait a sec. Couldn't I find what r equals is and then solve for the current ia? It would be : r(equal)= 24ohms. Voltage =50v so i=v/r would be i= 50/24 = 2.08.
Would this be allowed to find the current ia ? the answer in the book is 2 A.

9. Feb 3, 2010

### cepheid

Staff Emeritus
No, you can't apply this method as though the 20 ohm and 4 ohm were in series, because they aren't. The current is NOT the same in the 4 ohm resistor as it is in the 20 ohm resistor. That is why you cannot use your method. At node 1, the current that went through the 4 ohm resistor SPLITS into Ia and Ib. This is what I explained in my first post. Apply the method I suggested.

10. Feb 3, 2010

### vk6kro

You could work out the total resistance and get the current out of the power source from V / R.

Then you could work out the voltage across the 4 ohm resistor. (using I * R)

Subtract this from the supply voltage and you get the voltage across the parallel combination.

You now know the voltage across the 20 ohm resistor and could work out the current through it.
Similarly, you could work out the current through the 80 ohm resistor by the same method or by subtracting the 20 ohm current from the total current.

The total resistance isn't 24 ohms. The 20 ohms has 80 ohms in parallel with it.

11. Feb 4, 2010

### LP20

I understand it now!! finally, took a full day though. I got 20 ohms for total resistance, than I did find the rest from that, using kvl's. Thanks for helping me out!!