Ammeters problem. (DC Circuits)?

Question

I'm not sure how to solve problem #4 on the following Web Site:

http://www.physics.purdue.edu/~clarkt/Courses/Physics271/Exam%20Archive/Exams/P271F00FinalExam.pdf

Additional question:

The ammeter and the source of emf are now physically interchanged. Show that the ammeter reading remains unchanged. Assume the ammeter to have zero resistance.

Answer 1

Assuming a perfect ammeter:

R2 and R3 in combination make 4*6/10 = 2.4 ohms.

That added to R1 makes 4.4 ohms.

So total current is

I = E/R = 5/4.4 = 1.136 amps

This produces a voltage across R2 and R3 of

E = IR = 1.136*2.4 = 2.727 volts

That voltage across R3 produces a current of

I = E/R = 2.727/6 = 0.4545 amps

2)

R1, R2 in parallel are 2*4/6 = 1.333 ohms

in series with R3 that is 7.333 ohms

So total current is

I = E/R = 5/7.333 = 0.6818 amps

This produces a voltage across R2 and R1 of

E = IR = 0.6818*1.333 = 0.9090 volts

That voltage across R1 produces a current of

I = E/R = 0.9090/2 = 0.4545 amps

.

Answer 2

5/11 Amps;

Additional Q: 5/11 again

Answer 3

Solve the problem with current loops.  let the current in the loop with the battery be I1 and the current in the loop with the ammeter be I2.  Now sum voltage drops within each loop:

E + (I1 - I2)*R2 + I1*R1 = 0

(I2 - I1)*R2 + I2*R3 = 0

solve the two equations for I1 and I2.  The ammeter reads I2

After the interchange the equations read

(I1 - I2)*R2 + I1*R1 = 0

E + (I2 - I1)*R2 + I2*R3 = 0

do the same, except now the ammeter reads I1