Electrical – How to apply nodal analysis when value of dependent voltage source depends on current of a plain wire

circuit analysisvoltage-source

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Need to find out the value of V11 (V11 = 4*I9 Volt), as it's a dependent source and it's value is depending on the current through the plain wire W9 of no resistance. I solved this with Mesh current analysis method but I need to learn to be able to solve this using node voltage analysis method too.

Answers from mesh analysis currents and voltage drops of the elements.
I2 = 7 A & V2 = 280 V
I4 = 8 A & V4 = 40 V
I11 = 18 A & V11 = 40 V
I7 = 8 A & V7 = 80 V

My problem: While solving this with node analysis, as this problem belongs to supernode method application where we form 2 equations, one of which is given by difference of the two end nodes equals the voltage source for which I need value of I9 to be written in voltage form. Which I do not know how to form.

Update 2: I learned from some answers here that, it is possible to find the current and voltage values of each element without having to convert I9 to voltage form. Got the answers but they don't match the answers i got in mesh analysis method. Here's what I did.

Considering bottom node as reference and other nodes 'node a', 'node b' and 'node c' respectively from left to right and Vc=-240, I got these two equations..

node a + node c gives

(-19 + Va/40 + (Va-Vb)/5 + Iac) + ((Vc-Vb)/10 – Iac)= 0
=>-19 + Va/40 + (Va-Vb)/5 + Iac + (Vc-Vb)/10 – Iac = 0
=>0.025Va + 0.2Va-0.2Vb -24-0.1Vb = 19 since (Vc=-240)

0.225Va – 0.3Vb = 43 ——(1)

node b eqn.

(Vb-Va)/5 – 2*(Vb-Va)/5 + (Vb-Vc)/10 = 0
=>-(Vb-Va)/5 + (Vb-Vc)/10 = 0
=>-0.2Vb+0.2Va + 0.1Vb-0.1*-240 = 0

0.2Va – 0.1 Vb = -24 ——(2)

Giving me Va=307V and Vb=373V, giving voltage drop across R4=67V which is not the same as returned in mesh analysis method i.e. 40 V.

Please help me get it right or suggest me where I'm going wrong.

Thank you…

Best Answer

  • 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)