calculating indication of an ideal ammeter

Thread Starter

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.
 

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.,
 
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