# Calculating currents with a current source in series

Is this circuit correct? I am not sure if I can consider I5 as a current source. I remember that current sources cannot be in series.

Or is this as easy as it looks. Specifically I am getting I1=0.8232 I2=0.5488 I3 = 0.5875 I4 = 0.783 A. The confusion is caused by I5 which I believe is a current source. • Given the schematic, I5 should not be a current source (since no current source is drawn). However, if this circuit were just a 12V source and the 4 resistors drawn, I5 would be 1.37A, not 5.0A. Looks like my calculations match yours, so I'd guess this is just a bogus schematic/problem. – Shamtam Nov 26 '13 at 3:30
• If 5A is flowing as shown then you need to consider it part of the question if you are going to attempt to answer it (which you have done) OR give a decent argument why the 5A should not be considered part of the question and don't answer it. – Andy aka Nov 26 '13 at 8:28
• Well using nodal analysis by taking junctions A and B across I 5 gives a system of equations with no solutions. Specifically (0-Va)/6 + (0-Va)/9 = 5, 5 = (Vb-12)/12 + (Vb-12)/9, Va = Vb. – Dimo Nov 26 '13 at 14:20

## 1 Answer

There is no law forbidding current sources connected in series to other elements.

I5 must be generated by a current source to make the circuit work. Without a current source I5 is different from 5.0 A.

If I5 is replaced by a current source with I5=5.0 A, you get the following voltages: U1=U2=18V, U3=U4=(5*36/7)V=25.71V Which means, your imaginary current source I5 must provide a voltage of -31,71V. This is rather uncommon for real current sources, but not impossible to build.

Otherwise the 5.0A cannot be explained.

(all voltages named after current indices, voltage counting directions according to current directions)