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

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.

Thank you...

• First, specify a 0V reference - e.g. if it's the bottom rail, the voltage at W1/W10/R2/R4 node will be (-V8 - V11). Call this node A. You need one further node: R4/R7/I5, say, node B. These two nodes will be sufficient to solve. But, which direction is i9? It makes a difference.
– Chu
Mar 5 '17 at 9:55
• I4's direction matters, too.
– jonk
Mar 5 '17 at 16:07
• Assume the direction of both i4 and i9 to be right to left. Mar 5 '17 at 20:46
• @Chu I tried solving with just the two nodes and got the answers...which didn't match the results i got in the mesh analysis method. These were my equation... 0.225Va - 0.3Vb = 43 ------(1) 0.2Va - 0.1 Vb = -24 ------(2) resulting Va=307 and Vb=373 giving the voltage drop across R4=67V which is not the same as returned in mesh analysis method i.e. 40 V. Can you please help me with this where I am doing wrong? Mar 7 '17 at 13:39
• I wouldn't call a wire a "circuit element". A wire connects circuit elements (e.g. resistors, sources) and is part of a node.
– Curd
Mar 7 '17 at 14:24

• 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)

• Isn't Vc=-240? Since Vd=0; 0-Vc=240; -Vc=240 or Vc=-240. Mar 8 '17 at 14:54
• Yes, Vc= -240 if you choose Vd as references. Just write basic equations and then write reference node as a final equation (Vd = 0) and solve using matrix. Mar 8 '17 at 15:26
• @Darian lee: yes. I've corrected $V_c = -240V$
– Curd
Mar 8 '17 at 15:42
• Thanks @Curd, I proceeded as you suggested...Here's what I did...I11=19 + (Va-Vb)/5 - Va/40; & I9 = I11 - I7; Solving (Vc-Va) = 4*(I11 - I7) gives 0.425Va - 0.1Vb = -103 and other eqn for node b is 0.2Va - 0.1Vb = -24....Solving both I get Va=351. However it should be 280V...Did I do something wrong? Please help... Mar 8 '17 at 15:51
• @Darian lee: see my edit.
– Curd
Mar 8 '17 at 16:24

In nodal analysis you can deal with current sources by simply including them in the equations you write rather than a voltage divided by resistance between nodes.

So if you had Vx and Vy connected by Rz your nodal equation for Vx would contain one term (along with others for other nodes connected) that is:

(Vx-Vy)/Rz + other terms = 0

Now imagine you had the current source Iw coming out of Vx also then you would have:

(Vx-Vy)/Rz + Iw + other terms = 0

Now imagine that Vx was connected to Vu through dependant source Va=5*Iw

Your set of equations (assuming no other nodes or connections) may be:

(Vx-Vy)/Rz + Iw = 0

Vx-Vu = Va

Va = 5*Iw

NOTE: This is an example and does not represent a complete set of nodal equation, if it were there would be n equations and n unknowns (easily solved with matrices)

• @Wolfgand_Horton, Thanks for the reply, You're right but I need the actual value of V11 in terms of volts. In order to get form the equation we need all the terms in voltage form. It's already given to us that V11 = 4*I9 Volts.I need value of I9 to be written in voltage form. I need to know how to form the super-node equation (or Super-Super-Node in this case as 2 voltage sources are connected together) for the dependent voltage source when it is depending on the value of current through a wire of no resistance. Mar 5 '17 at 7:59
• Trust me if you write all of the equations in standard form following the procedure described above you don't need to do anything special... Nodal analysis is actually all about currents (Va-Vb)/R is equal to the currents through the branch. And if you have to nodes connected by voltage source Vc you can just write Va-Vb=Vc Mar 5 '17 at 8:06
• So write all of the equations as I described above including the equations that define the dependant sources. Solve them as a matrix and then you will get all values. If you need help solving systems of equations via matrix method I suggest looking into it for these types of problems. "Super node" as you describe it is unnecessary and is just another way to have equation Va-Vb = Vc.... In your scenario is would be Va-Vc = V11 and V11 = I9*4 Mar 5 '17 at 8:13
• Also don't think about "Wires of no resistance" that is just part of the node. If there is no component or source connecting to spots it is all one node... No need to think about a wire there. The labeling of W5, W3,W9 etc. has no impact on how you solve the problem and should be ignored completely when you write system of equations... Mar 5 '17 at 8:15
• I could write the node equation no problem, but what to do with the term I9 among all the voltages in the equation. How to eliminate the term I9 from the equation? Mar 5 '17 at 8:19

Here is an example of equations for node a:

-19A + (Va-Vd)/40 + (Va-Vb)/5 + Iac = 0

Va - Vc = -V11

V11 = 4*I9

(Note: I included Iac to represent the current flowing through source V11 (from a to c.. would be -Iac for node c equation), this will be one of your unknowns, but after writing all equations you should have tools to solve for this also).

Hope this helps to get started!

• And as for I5 = 2*i4... I am assuming that i4 is referring to the current through resistor 4.. So you could write: I5 = 2*(Va-Vb)/5 Mar 5 '17 at 8:39
• Here's what i tried but didn't get the answers as given to me by mesh analysis method.Please suggest me where I am wrong.. Assuming Vd=0 being reference node and Vc=-240 volt being connected to ground. Node c eqn. (Vc-Vb)/10 - Iac = 0 Adding node a and node c eqns gives,1st eqn for node a as given by you is -19 + Va/40 + (Va-Vb)/5 + Iac added to node c, being a part of supernode becomes... -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) Mar 6 '17 at 20:03
• 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 Gives me Va=307V and Vb=373V, giving voltage drop across R4=67V This is the answers from the mesh analysis method. currents and the voltage drops at the elements.. I2 = 7 & V2 = 280 I4 = 8 & V4 = 40 I11 = 18 & V11 = 40 I7 = 8 & V7 = 80 sorry for the bad formatting of the comment..line break is not working... Mar 6 '17 at 20:05
• Can you please suggest me where I am going wrong...That will be a huge help... Mar 8 '17 at 15:09
• Will look over your equations later.... But my first suggestion would be to not do so much algebra (easy to make mistakes). These equations are all linear and can be solved very efficiently on any graphing calculator with matrices Mar 8 '17 at 15:30