Okay here I go. I just started learning nodal analysis technique but I feel its difficult in solving the circuits consisting of dependent current and voltage sources.

Im unable to solve this circuit.enter image description here

The current source is a dependent current source and whose value is 180 times Ix and Ix is the current flowing through the 0.7 v battery.

Generally to solve this type of circuits we find a relation to express Ix in terms of node voltages / resistance. example Ix = (V2 - V1 ) / 500 ohms . But here there is no resistance in the 0.7 v branch voltage. I am unable to find a solution to solve it. Could anyone kindly explain me the solution to this circuit.

NOTE I have found a solution by finding the thevinin's resistance across 0.7 v battery and replaced it with Rth in series with 0.7v and I have got the solution but I want to know, Can we get it using only KCL equations (nodal analysis) ?

Please help me in solving this circuit and kindly explain it with the KCL equations. Your help is greatly appreciated. Thank you


3 Answers 3


AFAIK, you can always solve any linear circuit the 'brute force' way using nodal analysis:

  1. Write Kirchoff's Current Equations on all nodes except ground
  2. For every circuit component, (i.e. resistors, capacitors etc.), write down their behaviour (for instance, ohm's law for a resistance, i = c dV/dt for a capacitance and so on)
  3. At this point, we'll have a handful of equations with us. We can also try to eliminate as many equations from them as possible using any info we have; however in the end, we need to be left with N simultaneous equations in N unknowns. Solve them and we'll get all the node voltages and branch currents.

Coming to the circuit above, let's define the current through V2 as I2, and the ones through R_n as I_n. Let me also call the node at the top as V_a, the node between the CCCS and R5 as V_c, the one between the R_7 and R_8 as V_b and the node in the middle as V_e. Now, writing Kirchoff's Current Law on these nodes will leave us with $$ I_2 = I_7 + I_5\\ I_7 = I_8 + I_x\\ I_x + I_5 = I_6 $$ respectively. Writing down the 'behaviour' of R6, R5, R7, R8, V2, V3 and the CCCS will respectively yield $$ V_E = I_6 R_6 \\ V_A - V_C = I_5 R_5 \\ V_A - V_B = I_7 R_7 \\ V_B = I_8 R_8 \\ V_A = V_2 \\ V_B = V_E + 0.7\\ I_5 = 180 I_x $$

That's 10 linear equations in 10 unknowns. Solve them, and we'll find all I_x as 88.18uAmps...

Of course, 10 equations is a bit too much to solve by hand (I generally use Gauss-Jordan elimination to do this part), but as far as I've seen, this method works in situations where the usual 'text-book' approach using nodal and mesh analyses fail. Furthermore, we don't have to deal with the painful Thevenin equivalent/Super-mesh workarounds here...

On the downside, I'm not quite sure if this approach works with every possible circuit (so far I haven't seen any where it fails), so any negative feedback on this part is welcome :)

  • \$\begingroup\$ Thank you so much.. That is the right answer :) but could you explain me the equation Ve = I5 * R6 ? how did we get . Sorry but I find it difficult myself to calculate voltages like Vb or Va. Could you tell me any simple way to understand those things or provide some good links where I can improve my skills kindly explain me this please. Thank you @nav \$\endgroup\$
    – niko
    Commented Feb 20, 2013 at 18:15
  • \$\begingroup\$ I guess Ve = I6 * R6 ? \$\endgroup\$
    – niko
    Commented Feb 20, 2013 at 18:18
  • \$\begingroup\$ Please make it complete \$\endgroup\$
    – niko
    Commented Feb 20, 2013 at 19:33
  • \$\begingroup\$ Sorry for that typo; It's V_E = I_6 * R_6 as you said. Fixed :) \$\endgroup\$
    – nav
    Commented Feb 21, 2013 at 9:49
  • \$\begingroup\$ @niko: Frankly speaking, analog circuits are real sources of pain. To solve them, we really don't need anything more than what our textbooks provide, but then, we'll always stumble upon some exceptional circuit topology where the 'regular' textbook methods fail; however AFAIK, they can always be solved using the same methods, only that we should not restrict ourselves to using a single method (nodal/mesh). If you can figure out how a "(resistor + current_source in series) parallel to (volt_source + resistor in parallel)" works, I think you'll be able to play with any circuit. Good luck:) \$\endgroup\$
    – nav
    Commented Feb 21, 2013 at 10:51

The problem with solving this circuit by nodal analysis is not the CCCS, it's the voltage source V3. In nodal analsyis, voltage sources can't be directly accomodated but have to be accounted for bye creating "supernodes". In this case you'd combine the nodes at the left and right of V3 to make a supernode.

But of course that means that Ix is no longer a variable in the analysis.

Luckily, the basics of nodal analysis give a clue how to solve this. We just need to apply KCL at one of the nodes connected to V3.

For example, if we define the currents through R7 and R8 as I7 and I8, both defined in the "top-to-bottom" direction as the circuit is drawn, then we can easily find

Ix = I7 - I8

Since I7 and I8 can be written in terms of the node voltages, we can now continue with the nodal analysis.


In nodal analysis the unknowns to be solved for are the node voltages. In your circuit you have 4 nodes. The reference node is clearly known. The 20V voltage source reduces the unknowns by 1. The 0.7V source further reduces the number of unknowns:

V2=V1+0.7 (1)

As you have only one unknown (V1), you need to write one equation: KCL at node 1. The current Ix through the 0.7V source can be found in terms of the nodal voltages: Ix=(20-V2)/640-V2/100. Substitute V2 in terms of V1 (1) and you end up with one equation and one unknown V1.

The tableau analysis technique explained by @nav is a systematic technique to solve circuits, but it has many more equations. The modified nodal approach (MNA) is another systematic technique commonly used in commercial simulators.

In MNA, you would write KCL at each node (except ground) and consider the currents in the voltage sources as additional unknowns. To compensate the increase in unknowns you would add equations similar to (1).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.