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Find v and i in the circuit of Fig.3.11

I modified the circuit like:

and renamed the nodal voltages:

Now applying KCL at the supernode, $$i_1+i_2=3.5....(i)$$ We know, $$i_1=\frac{v_1}{\frac{12}{7}}, i_2=\frac{v_2}{\frac32} (\text{Note that }\frac{12}{7}=1.714)$$ From equation(i) we get $$\begin{align}\\ \frac{7v_1}{12}+\frac{2v_2}{3} &=3.5\\ \implies 7v_1+8v_2 &=42.......(ii)\\ \end{align}\\ $$ To get the relationship between \$v_1\$ and \$v_2\$ we apply KVL to the following circuit

$$-v_2+v_1+6=0\\ \implies v_2=v_1+6.....(iii)$$ Now, from equation (ii), $$7v_1+8(v_1+6)=42\\ \implies 15v_1=-6\\ \implies v_1=-0.4V$$ From equation (iii), $$v_2= -0.4+6=5.6V$$ So, \$i_1=-\frac{7*0.4}{12}=-0.233A\$. Hence, from the main figure \$v=-0.233*3=-0.699V\$. But, in my book the answer is \$v=-400mV\$. Please tell where is my mistake.

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  • \$\begingroup\$ First thing, your 3.5 A source is pointing the wrong way to be equivalent to the 14 V source it’s supposed to be replacing. \$\endgroup\$
    – The Photon
    Jul 29, 2020 at 5:52
  • \$\begingroup\$ Converting the 14 V source to a current source makes the problem more difficult. \$\endgroup\$
    – Chu
    Jul 29, 2020 at 6:16
  • \$\begingroup\$ @Chu please add an answer. I couldn't find any other choice. \$\endgroup\$
    – Ankita Pal
    Jul 29, 2020 at 6:25
  • \$\begingroup\$ What are the three currents flowing away from the \$v\$ node? \$\endgroup\$
    – Chu
    Jul 29, 2020 at 6:58
  • \$\begingroup\$ @Chu I applied the KCL in the 3rd figure. You can see there are three current around the supernode. \$\endgroup\$
    – Ankita Pal
    Jul 29, 2020 at 7:31

1 Answer 1

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Well, I am trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\text{I}_x+\text{I}_2\\ \\ \text{I}_x=\text{I}_3+\text{I}_4\\ \\ \text{I}_1=\text{I}_2+\text{I}_5\\ \\ \text{I}_5=\text{I}_3+\text{I}_4 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$

We also know that \$\text{V}_x=\text{V}_2-\text{V}_1\$.

Now, because you've to do the math to solve this. I will present a Mathematica code that will solve this problem:

In[1]:=FullSimplify[
 Solve[{I1 == Ix + I2, Ix == I3 + I4, I1 == I2 + I5, I5 == I3 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1)/R2, I3 == (V2)/R3, I4 == (V2)/R4, 
   Vx == V2 - V1}, {Ix, I1, I2, I3, I4, I5, V1, V2}]]

Out[1]={{Ix -> ((R3 + R4) (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I1 -> (R3 R4 Vi + R2 (R3 + R4) (Vi + Vx))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I2 -> (R3 R4 Vi - R1 (R3 + R4) Vx)/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I3 -> (R4 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I4 -> (R3 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I5 -> ((R3 + R4) (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  V1 -> (R2 R3 R4 Vi - R1 R2 (R3 + R4) Vx)/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  V2 -> (R3 R4 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)}}

Now, using your values we get:

In[2]:=R1 = 4;
R2 = 3;
R3 = 2;
R4 = 6;
Vi = 14;
Vx = 6;
FullSimplify[
 Solve[{I1 == Ix + I2, Ix == I3 + I4, I1 == I2 + I5, I5 == I3 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1)/R2, I3 == (V2)/R3, I4 == (V2)/R4, 
   Vx == V2 - V1}, {Ix, I1, I2, I3, I4, I5, V1, V2}]]

Out[2]={{Ix -> 56/15, I1 -> 18/5, I2 -> -(2/15), I3 -> 14/5, I4 -> 14/15, 
  I5 -> 56/15, V1 -> -(2/5), V2 -> 28/5}}

So, the answers are \$\text{v}=\text{V}_1=-\frac{2}{5}=-0.4\space\text{V}\$ and \$\text{i}=\text{I}_3=\frac{14}{5}=2.8\space\text{A}\$.

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  • \$\begingroup\$ You did it absolutely correct!!👍🏻👍🏻 \$\endgroup\$
    – Ankita Pal
    Jul 29, 2020 at 10:28
  • \$\begingroup\$ @AnkitaPal Thanks! \$\endgroup\$ Jul 29, 2020 at 10:43

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