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I'm unable to figure out how to use the current dependant voltage source in KVL

The following are the equations I've found, up till now: 20i+V2=10, V2=50i1, 20i+50i1=10, i=i1+i2

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    \$\begingroup\$ Use mesh analysis. Update your question with the equations you found and we will help you along. \$\endgroup\$
    – Huisman
    Feb 8 '20 at 12:19
  • \$\begingroup\$ updating your question means editing it, @Aimen: That way, others don't have to read all through the comments here. \$\endgroup\$ Feb 8 '20 at 12:22
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    \$\begingroup\$ You're on the right way. There is another way to write out V2, i.e. using the components between nodes b and c and between c and d. (BTW, you need to use 2 spaces at the end of a line to start text on a new line) \$\endgroup\$
    – Huisman
    Feb 8 '20 at 12:32
  • \$\begingroup\$ What happened to the 25 ohm resistor? It is not in your equations. (And no, in this circuit you can't ignore it) \$\endgroup\$
    – Oldfart
    Feb 8 '20 at 12:37
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    \$\begingroup\$ @Jan Yes I know, I got the right answers now. Thank you. \$\endgroup\$
    – Aimen
    Feb 8 '20 at 15:32
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Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

Using KVL, we can write:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{a}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1}{\text{R}_3}\\ \\ \text{V}_\text{x}=\text{V}_1-\text{V}_2=\alpha\text{I}_1 \end{cases}\tag2 $$

Using some Mathematica code:

FullSimplify[
 Solve[{I1 == I2 + I3, I1 == (Va - V1)/R1, I2 == V2/R2, I3 == V1/R3, 
   Vx == V1 - V2 == α*I1}, {I1, I2, I3, V1, V2, Vx}]]

Using Your values we get:

FullSimplify[
 Solve[{I1 == I2 + I3, I1 == (10 - V1)/20, I2 == V2/25, I3 == V1/50, 
   Vx == V1 - V2 == 5*I1}, {I1, I2, I3, V1, V2, Vx}]]

Which gives:

{{I1 -> 1/4, I2 -> 3/20, I3 -> 1/10, V1 -> 5, V2 -> 15/4, Vx -> 5/4}}

Which approximate to:

{{I1 -> 0.25, I2 -> 0.15, I3 -> 0.1, V1 -> 5., V2 -> 3.75, 
  Vx -> 1.25}}
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    \$\begingroup\$ Alright, I got it. Thank you very much for the help 😊 \$\endgroup\$
    – Aimen
    Feb 8 '20 at 13:10
  • \$\begingroup\$ @Aimen You're welcome, I'm glad that I could help. \$\endgroup\$
    – Jan
    Feb 8 '20 at 13:11
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    \$\begingroup\$ Please stop giving away homework solutions. \$\endgroup\$ Feb 8 '20 at 13:26

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