Find all node voltages in the circuit with respect to the bottom node as ground, where $R_1=100\Omega$, $R_2=5\Omega$, $R_3=200\Omega$, $R_4=50\Omega$, $V=50V$, $I=0.2A$. Use both node voltage and loop current methods.

Solution:

• Node voltage: Let the bottom node be ground and other node voltages be $V_1$ (left), $V_2$ (middle) and $V_3$ (right).
  
  $$\begin{array}{ll} \mbox{left node:} & V_1/200+(V_1-V_2)/5+(V... ...2=0 \\ \mbox{voltage source:} & V_3=V_2+V=V_2+50 \end{array}$$
  
  
  Solving this we get:
  
  $$V_1=-45.23,\;\;\;\;\;V_2=-48.69,\;\;\;\;V_3=1.31$$
  
  
  Alternatively, use the figure on the right and assume between the current and voltage sources is grounded $V_2=0$, then $V_3=50V$, and denote previous ground by $V_0$. We have
  
  $$\begin{array}{ll} \mbox{middle node $V_1$:} & V_1/5+(V_1-V_0... ...ottom node $V_0$:} & (V_0-V_1)/200+(V_0-50)/50=0.2 \end{array}$$
  
  
  Solving this we get:
  
  $$V_1=3.46,\;\;\;\;\;V_0=48.7,\;\;\;\;V_3=50,\;\;\;\;V_2=0$$
  
  
  Treating $V_0$ as ground, we get the same result as before:
  
  $$V_1=-45.2,\;\;\;\;\;V_0=0,\;\;\;\;V_3=1.3,\;\;\;\;V_2=-48.7$$
  
  

  We see that the second method is easier.

• Loop current: Let the loop currents be $I_a$ (top), $I_b$ (left) and $I_c$ (right).
  
  $$\begin{array}{ll} \mbox{top loop:}\;\;\;\;\;\;& 100I_a+50+5(... ...m loop:}\;\;\;\;\;\;& 200I_b+5(I_b-I_a)-50+50I_c=0 \end{array}$$
  
  
  Solving this we get:
  
  $$I_a=-0.465,\;\;\;\;\;I_b=0.226.\;\;\;\;\;\;I_c=0.026$$
  
  
  Alternatively, use the figure on the right and assume loop currents $I_a$ through voltage source $V$, $I_b=-0.2$ through current source and $I_c$ through $R_4$. We have
  
  $$\begin{array}{ll} \mbox{top loop: }\;\;\;\;\;\;\;&100(I_a-I_... ...:}\;\;\;\;\;\;\;&100(I_c-I_a)+50I_c+200(I_c+0.2)=0 \end{array}$$
  
  
  Solving this we get:
  
  $$I_a=0.49,\;\;\;I_b=-0.2,\;\;\;I_c=0.026$$
  
  
  Current through $R_1$ is $I_c-I_a=-0.464$, current through $R_3$ is $I_c+0.2=0.226$, same as before.

  We see that the second method is easier.