Find the current and voltage in the circuit

[diredragon]

Homework Statement

Homework Equations

3. The Attempt at a Solution [/B]
I have just started a course in electric circuits and it's rather simple but i don't get a few things. Mainly $U=IR$ and $E$ here is the voltage source. So i can use the voltage circulation rule to get the $I$ right?
$E-IR_2-IR_1-IR_3=0$
$I=\frac{E}{R_2+R_1+R_3}$
The voltage in the point $B$ is the $E-IR_2$ right? and the voltage in point $A$ must be zero because of the circulation. But what would be the voltage of the point between $B, A$. Should i do it like $E-IR_2-IR_1$ and to calculate it like this? Now this is a simple circuit but i don't get what happens if the current divides or if there are some more complicated schemes. Does this still apply? And is this right?

 

[NascentOxygen]

The voltage in the point B is the E−IR2E−IR2E-IR_2 right? and the voltage in point A must be zero because of the circulation. But what would be the voltage of the point between B and A.

voltage at B – voltage at A = voltage between B and A

Another way when you know the current through the resistors is to use Ohm's Law.

 

[diredragon]

voltage at B – voltage at A = voltage between B and A

Another way when you know the current through the resistors is to use Ohm's Law.

I would use ohm's law to find what if i know the current? The voltage? How would that work?

 

[cnh1995]

I would use ohm's law to find what if i know the current? The voltage? How would that work?

You can calculate the current with Ohm's law using the given data. What is the equivalent resistance of the three resistances? How are they connected?

 

[diredragon]

You can calculate the current with Ohm's law using the given data. What is the equivalent resistance of the three resistances? How are they connected?

$R_e=R_1+R_2+R_3$ and i get the same expression for $I$ using the ohm's law. But is there another way to calculate the voltage between point A and B? Using the way i described in post 1 i get $U_{ba}=10V$
I'll write it again.
$V_b=E-IR_2$
$V_a=E-IR_2-IR_1-IR_3$ right?

 

[cnh1995]

$R_e=R_1+R_2+R_3$ and i get the same expression for $I$ using the ohm's law. But is there another way to calculate the voltage between point A and B? Using the way i described in post 1 i get $U_{ba}=10V$
I'll write it again.
$V_b=E-IR_2$
$V_a=E-IR_2-IR_1-IR_3$ right?

Right. I believe this is the only convenient method.

 

[gneill]

Right. I believe this is the only convenient method.

One could take A as the reference node and write a node equation at B, solving for the potential there. But this is likely beyond the scope of what @diredragon has covered in circuit theory at this time.

 

[diredragon]

One could take A as the reference node and write a node equation at B, solving for the potential there. But this is likely beyond the scope of what @diredragon has covered in circuit theory at this time.

We talked about it a little bit and would like to understand it better by applying it to this circuit. Basically i take one node a make it a reference one where the potential is zero.
Basically we wrote something like this which i tried to understand: These are the set of equations that we wrote in order to apply the Node potential theory:
$1)$, $G_{11}V_1+G_{12}V_2+G_{13}V_3=I^{I}$
$2)$, $G_{21}V_1+G_{22}V_2+G_{33}V_3=I^{II}$
$3)$, $G_{31}V_1+G_{32}V_2+G_{33}V_3=I^{III}$
Where the G's are the $1/resistance$ and the numbers on them associate how we calculate them. I'm going to try this out on this problem:
Here i only have $G_{11}V_1=I^{I}$ here my reference node is $A$ and the node number 1 is $B$:
$G_{11}=\frac{1}{R_2}+\frac{1}{R_1}+\frac{1}{R_3}$
$I^{I}=\frac{E}{R_2}$
$\frac{R_1+R_2+R_3}{R_1R_2R_3}V_1=\frac{E}{R_2}$
Finally
$V_1=V_b=\frac{E}{\frac{R_1+R_2+R_3}{R_1R_3}}$
Aaand it's wrong isn't it xD? Does it even make sense?

 

[gneill]

You've almost got it. There's a small problem with your conductance term. You want $G_{11}$ to be the sum of the conductances attached to node 1 (that is, node A). There are two branches leading away from node 1, so you need to write separate conductances for each and sum them. Recognize that R1 and R3 are in series in one branch, and that conductances in series do not simply add.

 

[diredragon]

You've almost got it. There's a small problem with your conductance term. You want $G_{11}$ to be the sum of the conductances attached to node 1 (that is, node A). There are two branches leading away from node 1, so you need to write separate conductances for each and sum them. Recognize that R1 and R3 are in series in one branch, and that conductances in series do not simply add.

Do you mean node B? Couse A is my reference point and there the potential is 0?

 

[gneill]

Do you mean node B? Couse A is my reference point and there the potential is 0?

Yes, sorry, node B.

 

[diredragon]

You've almost got it. There's a small problem with your conductance term. You want $G_{11}$ to be the sum of the conductances attached to node 1 (that is, node A). There are two branches leading away from node 1, so you need to write separate conductances for each and sum them. Recognize that R1 and R3 are in series in one branch, and that conductances in series do not simply add.

I don't completely understand. Regarding the first three equations for node analysis i wrote. From there i needed only one couse i have two nodes and i set $V_2$ to be zero, right? So my $G_11$ is suppose to be the sum of conductances that lead from that node to another, in this case $R_1,R_2,R_3$ all led from node 1. The right side is suppose to be the current which in this case is only the voltage source divided by the only resistor that is on the wire that connects the source and node 1 right? So what was wrong? Is this thinking wrong?

 

[gneill]

The problem is with the treatment of R1 and R3. They are in series, so the effective resistance in that branch is R1 + R3. The conductance is therefore 1/(R1 + R3), not (1/R1) + (1/R3).

A quick technique for writing a node equation by inspection is to sum all the currents flowing out of a chosen node and set the sum equal to zero (You could also sum the currents flowing into the node, it's your choice). For each branch you take the potential difference across the branch (as determined by the node potentials at each end of the branch and taking into account any voltage sources in the branch) and divide by the net resistance in the branch. So for this circuit at node B:

$\frac{VB - E}{R2} + \frac{VB}{R1 + R3} = 0$

$\frac{VB - 12}{50} + \frac{VB}{100 + 150} = 0$

Solve for VB

 

[diredragon]

The problem is with the treatment of R1 and R3. They are in series, so the effective resistance in that branch is R1 + R3. The conductance is therefore 1/(R1 + R3), not (1/R1) + (1/R3).

A quick technique for writing a node equation by inspection is to sum all the currents flowing out of a chosen node and set the sum equal to zero (You could also sum the currents flowing into the node, it's your choice). For each branch you take the potential difference across the branch (as determined by the node potentials at each end of the branch and taking into account any voltage sources in the branch) and divide by the net resistance in the branch. So for this circuit at node B:

$\frac{VB - E}{R2} + \frac{VB}{R1 + R3} = 0$

$\frac{VB - 12}{50} + \frac{VB}{100 + 150} = 0$

Solve for VB

I might be asking some stupid questions here but on the left side its $V_b-12/R_2$ the current through branch 1 right? This current is going into the node? And why is it like this instead of $12-V_b$. And for the right branch its $V_b$, the potential difference part only couse there is no source on branch 2 right? And divided by the total resistance that makes the current that goes out from the node?

 

[gneill]

I might be asking some stupid questions here but on the left side its $V_b-12/R_2$ the current through branch 1 right? This current is going into the node? And why is it like this instead of $12-V_b$. And for the right branch its $V_b$, the potential difference part only couse there is no source on branch 2 right? And divided by the total resistance that makes the current that goes out from the node?

I chose to sum currents leaving node B. So the equation terms are written to reflect that choice. By summing all currents with the assumption that all are leaving (and none entering), the sum must be zero. It's a less error prone way to write KCL at a node if you assume all currents are flowing in the same direction (either all entering or all leaving). The sum will always be zero and you don't have to spend time trying to make educated guesses about the actual current directions. Just make the blanket decision that all current directions are the same and let the math sort it out automatically.

Your observations about the second term are correct. It could have been written as $\frac{VB - 0}{R1 + R2}$, where the "0" is the potential of the reference node, but it's convenient to just drop the "0" when writing the term.