# Find the voltage and the current in the circuit

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

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

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:

$$\z_1= \frac{-j}{2}\$$ $$\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)\$$.

• Voting to close as this appears to be a homework/test question with no work shown.
– vir
Commented Feb 1, 2023 at 21:05
• Welcome! Please show your work so far and where you are stuck. Commented Feb 1, 2023 at 21:08
• I added what I tried to do but I'm stuck. Commented Feb 1, 2023 at 21:41
• 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. Commented Feb 1, 2023 at 22:00
• The homework is open for any method ( so nodes or mesh method or circuit semplification ). Commented Feb 1, 2023 at 22:13

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.

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