0
\$\begingroup\$

I am trying to find out why this system:

enter image description here

should have an Open Circuit Voltage of:

\$U = U0 + I0*R2 + 0*R3\$

\$ = U0 + I0R2\$

I understand why the R3 component doesn't change anything but I fail to understand why R1 isn't contributing anything.

To be clear with all of you. This is an old exam question that I'm using to study for the test but can't find the answer in either the textbook or in the lecture material.

Thanks in advance

\$\endgroup\$
  • \$\begingroup\$ R1 does develop a voltage across its terminals, but it does not show on Vab. You can think that the contribute of R1 is to raise the potential on its left side. \$\endgroup\$ – Sredni Vashtar Oct 18 '16 at 23:40
  • \$\begingroup\$ "why R1 isn't contributing anything" why you think it should contribute?! \$\endgroup\$ – CroCo Oct 19 '16 at 0:38
3
\$\begingroup\$

Start with KVL,

enter image description here

-U0 - I0R2 + 0*R3 + U = 0

U = U0 + I0R2

\$\endgroup\$
1
\$\begingroup\$

Look at the current source and where the current must flow. It goes through \$R_1\$ and \$R_2\$, right? But when looking at the terminals \$A\$ and \$B\$, they only see the voltage of \$U_0\$ plus the drop across \$R_2\$. The drop across \$R_1\$ exists, but it doesn't have an impact on the terminals. It just impacts the potential difference across the current source. And terminals \$A\$ and \$B\$ don't care about that part of it.

Just walk through by summing the voltages around the loop from \$B\$ to \$A\$. No \$R_1\$ in there, right?

\$\endgroup\$
0
\$\begingroup\$
  • Short out the part with no current flowing ( Open circuit ) like R3

  • Short out the R values in series with a current source like R1 because it has no effect on the voltage drop on R2 in that single loop.

  • now what do you get?
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.