# Grounded electric circuit analysis

1. Nov 3, 2015

### Gbox

1. The problem statement, all variables and given/known data
Find the voltage in A,B,C,D when the circuit is grounded at point A

2. Relevant equations
$V=IR$
$\sum_{i=1}^{n}I_n=0$
$\sum_{i=1}^{n}v_n=0$
3. The attempt at a solution
I have managed to divide the current according to KCL, but I am not sure how should I approach building the voltage equations when the circuit is grounded, the sum of the voltage until A is zero? even if it is not a closed circuit?

2. Nov 3, 2015

### Staff: Mentor

"Grounding" the circuit at point A means that you take Node A to be the reference node for "measuring" potentials at other points in the circuit. Imagine that you have a voltmeter and place its negative lead on node A, then move the positive lead around to various places in the circuit to measure potentials.

You can analyze the circuit in any way you wish, determining potentials and currents as usual. Then do a "KVL walk" from A to any other place in the circuit, summing the potential changes along the way to find the potential at that place with respect to A.

How you go about analyzing the circuit is up to you, and will depend upon the methods that you've studied so far.

3. Nov 3, 2015

### Mister T

In other words, grounding at A means the value of the potential $V_A$ at A is set to zero. Thus, when they ask for the potential $V_C$ at C, what they are asking for is the potential difference $V_C-V_A$.

And another thing, I would suggest you review the guidelines for the correct number of junction equations in relation to the number of loop equations. You have 6 unknowns, so you need six equations. But they have to be independent equations, and choosing the right number of each type is essential to making that happen.

4. Nov 4, 2015

### Gbox

Can you please refer me to those guidelines? Will it be sufficient to create only 3 equations?

5. Nov 4, 2015

### Staff: Mentor

A straightforward approach, if you are using fundamental KVL and KCL equations is, when you are labeling the circuit for currents, don't just create new currents for every branch. When you come to a junction where a current splits, choose the fewest new currents you can get away with. So if for example you came to a junction with i1 flowing into it and two paths leading out, create "i2" on one of those paths and put "i1 - i2" on the other. If already created currents meet at a junction (flowing into it), what flows out must be the sum of them. You'll find that you create the minimum number of currents that way. So for example, beginning with i1 as the first "created" current:

Notice that at node C there are two new paths, and a new current i2 was created on one of them, the downward leg through the 112 V battery (although the choice is arbitrary), and so i1 - i2 is left for the other path. At node D a new current i3 is created for one path and the remainder i2 - i3 gets assigned to the other path. At B the (i2 - i3) and (i1 - i2) join together resulting their sum i1 - i3 leaving the node.

Now when you write your KVL loop equations you write as many equations as there are different currents. In this case you need three loop equations because there are three independent currents. Every component in the circuit should be included in at least one loop.