next up previous
Next: About this document ...

Example 4: (Homework) Find all node voltages with respect to the top-left corner treated as reference node:

currentmethod1.gif

$V_1=12 V$, $V_2=6 V$, $R_1=3 \Omega$, $R_2=8 \Omega$, $R_3=6 \Omega$, $R_4=4\Omega$.

Solution: For convenience, we first assume the bottom node $V_0=0V$ is reference point, and we have node voltages $V_1=12 V$ (left), $V_3$ (middle) and $V_2=-6V$ (right). Applying KCL to $V_3$, we get

\begin{displaymath}
\frac{V_3-(-6)}{6}+\frac{V_3-12}{3}+\frac{V_3}{8}=0
\end{displaymath}

Solving this equation we get $V_3=24/5$. Now treating $V_1$ as reference point (ground), we subtract $V_1=12$ from all node voltages to get $V_1=12-12=0$, $V_3=24/5-12=-36/5$, $V_2=-6-12=-18$, $V_0=0-12=-12$.

When the middle node is treated as ground, we can still solve the problem the same way as above by treating the bottom node is the ground, and then subtract $V_3=24/5$ from each of the voltages to get the answer.





Ruye Wang 2019-02-04