# calculating indication of an ideal ammeter

#### kawafis44

Joined Mar 13, 2008
2
that's the exercise

and my solution

calculations for A,B,C (1.1)
$$-I_1 +I_3 + I_5 = 0$$
$$-I_3 + I_4 - I_6 = 0$$
$$I_2 -I_4 -I_6 = 0$$

calculations for I, II, III (2.1)
$$-U_1 -U_3 +0 = 0$$
$$-U_2 -U_4 -0 = 0$$
$$U_3 + U_4 + 25 = 0$$
because $$U_6=0, E_5=25V$$

i use resistors $$R_1,R_2,R_3,R_4$$ in this formula $$U_i=R_iI_i$$
and i have got (2.2)
$$-2I_1 -8I_3 = 0$$
$$-1I_2 -9I_4 = 0$$
$$8I_3 + 9I_4 + 25 = 0$$

i try to calculate from 1.1 and 1.2 this current $$I_6$$ (i thinks this is what i need to find), but i cannot obtain the result. by the way calculations are too complicated for an exercise like this

greetings!

#### beenthere

Joined Apr 20, 2004
15,819
The ideal ammeter looks like a short circuit (or another length of wire). That means the current will only flow in two of the resistors. Ohm's law will let you figure the current easily.

#### Ron H

Joined Apr 14, 2005
7,014
That means the current will only flow in two of the resistors.
Come again

#### Ron H

Joined Apr 14, 2005
7,014
Here's a big hint. I did it using Thevenin's theorem.

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#### beenthere

Joined Apr 20, 2004
15,819
Probably too rough-and-ready of me, but I looked at the ideal ammeter as a short. That made the current path into the 8 ohm resistor, through the ammeter, and out the 1 ohm resistor. Solve for the pairs in parallel to get the voltage drops, and there is the current through those resistors (the 8 & 1 ohm) and therefore the ammeter.

#### Ron H

Joined Apr 14, 2005
7,014
Probably too rough-and-ready of me, but I looked at the ideal ammeter as a short. That made the current path into the 8 ohm resistor, through the ammeter, and out the 1 ohm resistor. Solve for the pairs in parallel to get the voltage drops, and there is the current through those resistors (the 8 & 1 ohm) and therefore the ammeter.
I'm not following you. Does that account for the current through the 2 ohm and 9 ohm resistors?

#### beenthere

Joined Apr 20, 2004
15,819
No. And on second look, my visualization doesn't look quite so correct. I saw the 8 ohm and 1 ohm as the current path through the ammeter.

#### hgmjr

Joined Jan 28, 2005
9,029
There are a number of ways to solve your problem but RonH's suggestion of using Thevenin's Theorem appears to represent a straightfoward solution.

One thing RonH, did you purposefully omit tying the two grounds together in your diagram.

I have altered you redrawn version of the schematic to show what I am referring to.

hgmjr

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#### Ron H

Joined Apr 14, 2005
7,014
There are a number of ways to solve your problem but RonH's suggestion of using Thevenin's Theorem appears to represent a straightfoward solution.

One thing RonH, did you purposefully omit tying the two grounds together in your diagram.

I have altered you redrawn version of the schematic to show what I am referring to.

hgmjr
Oops!
I forgot to draw that in. Good catch!
Obviously, no current will flow without a return path.

#### hgmjr

Joined Jan 28, 2005
9,029
I suspect that you had envisioned ground symbols drawn on each of the loops. I had to look at it a couple of time before I noticed there were no ground symbols. After you have looked at as many schematics as you and I have over the years, you tend to overlook a missing explicit connection and assume that they are connected together implicitly.

hgmjr

#### Ron H

Joined Apr 14, 2005
7,014
I suspect that you had envisioned ground symbols drawn on each of the loops. I had to look at it a couple of time before I noticed there were no ground symbols. After you have looked at as many schematics as you and I have over the years, you tend to overlook a missing explicit connection and assume that they are connected together implicitly.

hgmjr
Yeah, I have to confess that after I did the math, I checked my results with LTspice. LTspice won't let you run a sim without a GND. Before I posted, I took all of them out because the OP's schematic didn't have one, and forgot to put in the wire to complete the circuit.
I didn't actually simulate the intermediate step, but the copy and paste still had GND, which I removed.,