Electrical – Finding voltage drop on a resistance using mesh analysis

You don't need meshes. Just calculate everything one source at a time, while keeping the other sources zeroed out (i.e. a voltage source is replaced with a wire, and a current source is replaced with an open (cut) circuit). Then sum the results from each source.

So in this case, the voltage source would see nothing but four resistors in series, and both current sources would see (different sets of) 2 series resistors connected in parallel with 2 other series resistors. Hopefully this helps you see how to "zero" the sources.

I'm not familiar with MIT's Circuit Sandbox, but it shouldn't matter where you put the ground since you just need to check the current flowing in different places - so just pick a spot and call it ground!

• GND node is node d, i.e. \$v_d=0\$.
• Node a, c and node d form the supernode a-c-d.
• Set up nodal equations (KCL) only for node b like you have learn in standard nodal analysis (→ 1st equation)
• With super node method normaly you also get one KCL equation per supernode; but not in this case, because the supernode contains the GND node (remember: in nodal analysis you don't write a KCL equation for GND node); so also no KCL equation for supernode a-c-d here.
• But with supernode method you still you have to write for each voltage source that caused a node to be combined into a supernode one additional equation that expresses the potential difference between the nodes connected by the voltage source. In this case:
  • for voltage source V8:
    \$v_d - v_c = V_8 = 240V\$.
    Because \$v_d=0\$ this can be simplified to \$v_c = -240V\$
    (→ 2nd equation).
  • and for voltage source V11:
    \$v_c - v_a = V_{11} = 4i_9\$
    (→ 3rd equation).
• Express all occurences of currents (i.e. here \$i_9\$) by expressions of voltages and constant currents (=given values of independent current sources).
  And this step is probably what makes this problem a little bit tricky: \$i_9\$ and \$i_{11}\$ can not immediately be expressed by voltages because there are no resistors; but \$i_9\$ can be expressed as sum of \$i_7\$ and \$i_{11}\$ (watch for correct sign) and \$i_{11}\$ as sum of \$i_0\$, \$i_2\$ and \$i_4\$. So eventually all currents can be expressed by expressions of voltages and given current values.

Finally you get a system of 3 equation with 3 unknowns \$v_a, v_b\$ and \$v_c\$.

EDIT:
The equations are:

• KCL for node B:
  \$(v_b - v_a)/R_4 + (v_b - v_c)/R_7 = 2i_4\$
• Equation for voltage source \$V_8\$:
  ... \$v_c = -240V\$
• Equation for voltage source \$V_{11}\$:
  \$v_c - v_a = V_{11} = 4i_9\$

Remaining \$i\$'s that need to be replaced by expressions of \$v\$'s are:

• \$i_4 = (v_b - v_a) / R_4\$
• \$i_9 = i_7 + i_{11}\$
• \$i_{11} = i_0 + i_2 + i_4\$
• \$i_0 = I_0 = const\$
• \$i_2 = -v_a/R_2\$
• \$i_7 = (v_b - v_c)/R_7\$

If you substitute these \$i\$-identities you should get 3 equations that contain as unknowns only \$v_a\$, \$v_b\$ and \$v_c\$.

(Maybe there are some more little sign errors left for the attentive reader to find and correct)