cant anyone help me with measuring the voltage drop across a series-parallel circuit? R3 and R4 to be specific, i tried it and my math is not right some how, Vr4=(total current) * r4 but not getting the right answer

Question

briensmarandache · Answer

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anonymous · Answer

i get 0.721 mA for the current going trough R3 and R4, which gives a voltage drop of 0.238V for R4 and 1.08V for R3, adding those up i get: 1.32V for the total voltage drop across R3 and R4.
is that what you are trying to calculate?

michele_laino · Answer

I got these values:
current which flows through R_1: I=1 mA
voltage drop across R_1: V_1= 680* 10^(-3)=0.680 volts
voltage drop across R_2 and R_3: V_3-4= 2 -0.680=1.32 volts

briensmarandache · Answer

@perl can you help with this, not sure what values to use to get the voltage drop across R3

radar · Answer

Why can't you use the data provided by the measuring devices that are depicted in the diagram.  For example U1 indicates a value of .721 milliamp, using that value calculate the voltage drop across R3.  Er3 = 1.5kOhm* .721 ma. = 1.08 volts
U3 is providing the value for the voltage drop across R4 as .238 volts or (.721 ma. * 330 Ohm).  

Show your math and let us see where you are going astray.

anonymous · Answer

if you dont get the same values as we do, which are the same you got using Multisim, please you us your math, so we can find any mistakes you made there.