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Find \$v_2(t)\$ and \$i_5(t)\$

Data: \$i_4(t)= 3\cos(2t+150^\circ)\$

enter image description here

Solution:

\$v_2(t)= 68.807\cos(2t+7^\circ)~\mathrm{mV}\$

\$i_5(t)= 2.481\cos(2t-177^\circ)~\mathrm{A}\$

What I need is the calculation to find the solution.

Here is my attempt:

enter image description here

\$z_1= \frac{-j}{2}\$ \$z_3= \frac{-j}{6}\$

Nodes method:

$$\left[\begin{matrix} \frac{1}{\frac{-j}{2}}+1+\frac{1}{3-\frac{j}{6}} \\\\ \end{matrix}\right] \left[\begin{matrix} V_a \\\\ \end{matrix}\right] = \left[\begin{matrix} 2.598+1.5j \\\\ \end{matrix}\right]$$

This solves to:

\$V_a=0.957-1.424j\$

But from here I don't know how to find \$v_2(t)\$ and \$i_5(t)\$.

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  • \$\begingroup\$ Voting to close as this appears to be a homework/test question with no work shown. \$\endgroup\$
    – vir
    Feb 1, 2023 at 21:05
  • \$\begingroup\$ Welcome! Please show your work so far and where you are stuck. \$\endgroup\$
    – winny
    Feb 1, 2023 at 21:08
  • \$\begingroup\$ I added what I tried to do but I'm stuck. \$\endgroup\$ Feb 1, 2023 at 21:41
  • \$\begingroup\$ This is for a unit on nodal analysis and not another form of analysis correct? I ask because your required work would have different steps depending on the method. \$\endgroup\$ Feb 1, 2023 at 22:00
  • \$\begingroup\$ The homework is open for any method ( so nodes or mesh method or circuit semplification ). \$\endgroup\$ Feb 1, 2023 at 22:13

1 Answer 1

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Keeping \$\phi=150^\circ\$, KCL is:

$$\frac{V}{1\:\Omega}+\frac{V}{3+\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 3\:\text{F}}}+\frac{V}{\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 1\:\text{F}}}+3\exp\left(j 150^\circ\right)\:\text{A}=0\:\text{A}$$

This solves out to \$V=0.0741541291509657 - j1.23821041275527\$ or \$V\approx 1.24\:\text{V}\:\angle -86.57^\circ\$. Divide that by \$\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 1\:\text{F}}\$ to get the current of \$\approx 2.4809\:\text{A}\:\angle +3.43^\circ\$.

But since that calculation assumed the current opposite to the arrow, subtract \$180^\circ\$ from the angle to get the current in the indicated direction, or \$\approx 2.4809\:\text{A}\:\angle -176.57^\circ\$.

The final answer for \$i_5\$ is \$2.4809\:\text{A}\cdot\cos\left(2t-176.57^\circ\right)\$.

You could also assume the current source starts at \$\phi=0^\circ\$ (which is fine to do.) Then the KCL is:

$$\frac{V}{1\:\Omega}+\frac{V}{3+\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 3\:\text{F}}}+\frac{V}{\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 1\:\text{F}}}+3\exp\left(j 0^\circ\right)\:\text{A}=0\:\text{A}$$

This solves out to \$V=-\frac{1299}{1901}+j\frac{1968}{1901}\$ or \$V\approx 1.24\:\text{V}\:\angle 123.43^\circ\$. Divide that by \$\frac1{j\cdot 2\:\frac{\text{rad}}{\text{s}}\cdot 1\:\text{F}}\$ to get the current (in the direction opposite to the arrow.) I get \$\approx 2.4809\:\text{A}\:\angle -146.57^\circ\$. Add \$180^\circ\$ to get it in line with the indicated arrow, so \$\approx 2.4809\:\text{A}\:\angle +33.43^\circ\$

Since in this case I chose to start with \$\phi=0^\circ\$, the computed angle above is "with respect to" the current source phase. So here, I must add the angle back to the original source phase to find \$150^\circ+33.43^\circ=183.43^\circ=-176.57^\circ\$.

So, again, the final answer I get for \$i_5\$ is \$2.4809\:\text{A}\cdot\cos\left(2t-176.57^\circ\right)\$.

Done two different ways. But the same answer.

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  • \$\begingroup\$ Where did you get the 3A in the KCL? \$\endgroup\$ Feb 2, 2023 at 10:11
  • \$\begingroup\$ @user3576105 I edited the answer to clarify. Feel free to ask additional questions, if any. \$\endgroup\$ Feb 2, 2023 at 21:03

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