Help with this Physics Problem about circuits and current?

Drawing is too small. When I enlarge it, I still can't make out the subscripts for the 4 resistors.

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only way I can see to solve this is to make two thevenin equivalents, one with the left 2 resistors and battery, the other with the right two and a second 9v battery.

for the left, Rth = 2•6/8 = 3/2 ohms, Vth = 9(2/8) = 9/4 volts

for the right, Rth = 1/2 ohms, Vth = 9/2 volts

Combining, we have 3/2+1/2 = 2 ohms, and V = difference, or (9/2–9/4) = 9/4 volts

current is (9/4) / 2 = 9/8 amps

Simplify the resistors. The 2 above wire A in the drawing are in parallel. The equivalent resistance is

Req1 = R2*R1 / (R2+R1) = 12/8 Ohms = 1.5 Ohms

The 2 below wire A are also in parallel.

Req2 = R1*R1 / (R1+R1) = 4/4 Ohms = 1 Ohms

The 2 equivalent resistances, Req1 & Req2, are in series. So the circuit's total resistance is 2.5 Ohms.

So the total current is 9 V / 2.5 Ohms = 3.6 amps

So the voltage at the tie point of wire A is the voltage drop across Req2 = 3.6 amps*1 Ohm = 3.6 V. So the voltage drop across R2 is 9 V - 3.6 V = 5.4 V

So the current thru R2 is 5.4 V / 6 Ohms = 0.9 amps.

The total circuit current is 3.6 amps and, since the 2 resistors below wire A are equal, 1/2 of that current, or 1.8 amps, will flow thru each of them. Since only 0.9 amps goes thru R2, if the resistor at the bottom left is to have a current of 1.8 amps, 0.9 amps must have come right to left thru wire A and added to the 0.9 coming down thru R2.

I agree with your 2nd answer. If your instructor says 0.5, perhaps you should argue about that.

R2 = 3*R1 and the ameter has no resistance so the R2->R1 mid junction is at the same potential as the R1->R1 mid junction this means that three times as much current passes throughthe upper R1 a through R2. Then the two lower R1 resistors carry each carry half of the total current.

so call I2 the current through R2, then total current =4*I2. and the current through the ameter must therefore be I2. Now all you need to do is work out what I2 must be.

E = 6*I2+2*2*I2 = 9V so I2 = 9/10 Amperes

What? half an amp - how? If there's half an amp flowing through there then??? Thanks for posting - I'll have to have another look :( obviously

Had another look - it is 0.9A