Calculater voltmeter reading with internal resistance

[lumdef]

The task and my attempt to solve it are on the picture. The book gives 72.8 mA; 34.3 mA, 1.93 V as an answer. I am not sure I understood the task, and therefore drew the circuit correctly. I tried to do mesh analysis, but it does not yield the correct results.

[Kurets]

It seems you are drawing the circuit correctly. I will ignore the 1Ohm in my explanation.

The middle cell has so low impedance that it will completely dominate the voltage across the voltmeter. The voltage across 50ohm is therefore approximately 2V and you get 40mA to the meter. In the left part of the circuit you have 3.5V applied across the 100ohm resistor which results in 35mA though it.

So you will have approximately 40+35mA through the 2V cell, and 35mA through the 1.5V cell.

These numbers are not exact, as they ignored the 1Ohm output resistance of the 2V cell, but it shows that the answer from the book is correct.

[radiolistener]

U1 - I1*R1 - (I1+I2)*R3 = 0 => (R1+R3)*I1 + R3*I2 - U1 = 0 => 150*I1 + 50*I2 + 1.5 = 0
U2 - I2*R2 - (I1+I2)*R3 = 0 => R3*I1 + (R3+R2)*I2 - U2 = 0 => 50*I1 + 51*I2 - 2 = 0

150*I1 + 50*I2 + 1.5 = 0 => I2 = -3*I1 - 0.03
50*I1 + 51*(-3*I1 - 0.03) - 2 = 0 => I1=3.53/-103=−0.0343 A
I2 = -3*(3.53/-103) - 0.03 = 0.0728 A

I3 = I1 + I2 = −0.0343 + 0.0728 = 0.0385 A
U3 = I3*R3 = 0.0385*50 = 1.925 V

[lumdef]

My problem was that I wrote KVL equations incorrectly, here is the right way...