# How can I find the steady state of current I with sinusoidal voltage source?

I am reading the book Electric Circuits 10th Edition from Nilsson and Riedel. In page 321 has a problem that seems easy but I find a difficulty to solve it.The problem states:

A 90 ohm resistor, a 32 mH inductor, and a 5 mF capacitor are connected in series across the terminals of a sinusoidal voltage source. The steady-state expression for the source voltage v_s is 125 angle -60 degrees Volt and $$\\omega= 5000 rad/s\$$. Find the value of capacitance that yields a steady-state output current i with a phase angle of -105 degrees.

My take:

To find the value of capacitance that yields a steady-state output current with a phase angle of -105 degrees, i can use the impedance method for analyzing the circuit. In a series RLC circuit like this, the total impedance is the phasor sum of the individual impedances for the resistor (R), inductor (L), and capacitor (C).

The impedance (Z) of a resistor : $$\Z_R = R\$$

The impedance (Z) of an inductor : $$\Z_L = j \omega L\$$

where j is the imaginary unit, $$\\omega\$$ is the angular frequency (in radians per second), and L is the inductance.

The impedance (Z) of a capacitor is:$$\Z_C = 1 / (j\omega C)\$$

Now, i can calculate the total impedance (Z_total) of the series RLC circuit by adding the individual impedances:

$$\Z_{total} = Z_R + Z_L + Z_C\$$

Given:

$$\R = 90\$$ ohms

$$\L = 32 mH = 0.032 H\$$ (converted to henrys)

$$\\omega = 5000 rad/s\$$

$$\v_s = 125 \angle -60 degrees V\$$

First, let's calculate the impedances for each component:

$$\Z_R = 90\$$ ohms

$$\Z_L = j * 5000 rad/s * 0.032 H = j * 160\$$ ohms

$$\Z_C = 1 / (j * 5000 rad/s * 5 mF) = -j * 40\$$ ohms

Now, I calculate the total impedance:

$$\Z_{total} = Z_R + Z_L + Z_C\$$ $$\Z_{total} = 90 ohms + j * 160 ohms - j * 40\$$ ohms

Now, we have the total impedance in complex form. To find the current phasor (I), we can use Ohm's law:

$$\I = V_s / Z_total\$$

where V_s is the source voltage phasor:

$$\V_s = 125 \angle -60\$$ V

Now $$\I = \frac{125 \angle}{|Z| \angle \theta_z} = \frac{125 \angle -60}{|Z|}\angle -60-\theta_z\$$

Now In the solutions manual says that $$\ -60 - \theta_{z} = -105 \$$.What is the type that he used?

In addition they continue with :

$$\Z = 90 +j160+jX_c \$$

and they find $$\X_c= -70\$$ ohm or $$\X_c = -\frac{1}{\omega C}=-70\$$.How they found -70 ? Any help ? The capacitance is $$\ 2.86 \mu F\$$.

• Since your Vin is at - 60 degrees we can "shift it back" to 0 degrees by adding +60. Thus, the current phase shift needs to be -105+60 = -45 degrees. And the pahase angel is tg(-45) = -1. Thus, --- > (XL - XC)/R = |1| From this we can see that XC must be equal to 160 - 90 = 70 to get (XL - XC)/R = |1| ---> (160 - 70)/90 = 90/90 = |1|
– G36
Commented Sep 10, 2023 at 12:00
• Homer, let me re-phrase what G36 is saying. You know that the current is -45 degrees relative to the voltage. You know the real part is 90 Ohms. You know that one piece of the imaginary part is 160j. (The 5000*32mH.) But you need the imaginary part to be the same magnitude as the real part, because then you get an angle of 45 degrees (opposite side is equal to the adjacent side.) So you want a capacitor that is -70j so that 160j-70j=90j. That's all there is. After that, it's just turning that into a capacitor value. Commented Sep 10, 2023 at 12:40
• "Steady state" means that some derivative has gone to zero. I suppose that derivative could be that of the phase change, but I suggest it's lousy notation. Commented Sep 10, 2023 at 13:07
• @ScottSeidman Perhaps "equilibrium state" may be better? Commented Sep 10, 2023 at 13:25
• Equilibrium has an even stricter definition. I've seen this called the "natural response", as opposed to the "forced response". Commented Sep 10, 2023 at 13:56

I'm looking at the book problem in the 9th edition right now:

What they want you to think about is $$\I=\frac{125\:\angle -60^\circ}{\mid Z\mid\:\angle \theta}=\frac{125}{\mid Z\mid}\:\angle\left(-60^\circ-\theta\right)=\frac{125}{\mid Z\mid}\:\angle -105^\circ\$$.

From $$\-60^\circ-\theta=-105^\circ\$$ you should be able to find that $$\\theta=45^\circ\$$, quite easily.

Also, you know that $$\Z=90+j\left(5000\cdot 32\:\text{mH}\right)+jX_C=90+j160+jX_C\$$. Since to get $$\\theta=45^\circ\$$ you must have $$\Z=90+j90\$$ (the opposite side must be the same as the adjacent side of the triangle here), it follows that since $$\j\left(160+X_C\right)=j90\$$ then $$\X_C=-70\:\Omega\$$ and that given $$\X_C=-\frac1{\omega\:C}=-70\:\Omega\$$ then $$\C=\frac1{500\:\cdot\:70}\approx 2.857\:\mu\text{F}\$$.

That's part (a).

For part (b) it is then just falling off the log: $$\I=\frac{125\:\angle -60^\circ}{90+j90}\$$. But the magnitude of the divisor is just $$\\sqrt{90^2+90^2}\$$, so it follows that the current magnitude is then $$\\mid\: I\mid=\frac{125}{\sqrt{90^2+90^2}}\approx 982.1\:\text{mA}\$$.

So the final answer for (b) is, in polar, $$\I=982.1\:\text{mA}\:\angle-105^\circ\$$.

Rounding to three significant places gets you the book answer.

Footnote: "the opposite side must be the same as the adjacent side of the triangle here": It's just the use of the tangent function. There is both a geometry approach (triangles as pictures) and a trigonometric approach (algebra that relates all the picture details with expressions and equations.) You need to look that up. It turns out that $$\\tan\left(45^\circ\right)=1\$$. So this means that $$\y=x\$$:

This is basic trig/geometry. If you are not comfortable with it, you need to spend some time here.

(No Euler's required. Though in the long run it would also serve you well to understand how to apply Euler's. It's a brilliant piece of knowledge to own.)

• "Since to get $\theta=45^\circ$ you must have $Z=90+j90$ (the opposite side must be the same as the adjacent side of the triangle here), it follows that since $j\left(160+X_C\right)=j90$ then $X_C=-70\:\Omega$ " can you show me the math?It will be very helpful.Thanks in advance and your for answer.I appreciate it a lot Commented Sep 10, 2023 at 14:15
• @HomerJaySimpson I've added an explanatory section as a footnote. Hopefully, that helps out. Commented Sep 10, 2023 at 14:36
• That was very helpful.Thank you once again Commented Sep 10, 2023 at 15:39
• In my personal opinion, this book does not explain the concept very well (if it explains anything at all). This is why the OP has a problem with this.
– G36
Commented Sep 10, 2023 at 18:20
• @G36 do you recommend any other book that explains these concepts ? Commented Sep 10, 2023 at 18:51

You lose me a bit with the notation convention you're using in the line that start with "Now I=..."

I believe if we use exponential form for the phasors ($$\Ae^{j \theta}\$$), everything will be a lot clearer:

You are correct that we should apply the complex form of ohm's law:

$$\I = \frac{V}{Z} =\frac{A_v e^{j \theta_v}}{A_z e^{j \theta_z}} = \frac{A_v}{A_z} e^{j (\theta_v - \theta_z)} \$$

If the output phasor angle should be -105, then clearly: $$\-105 = \theta_v - \theta_z = -60 - \theta_z\$$. So $$\\theta_z = 45\$$.

-70 is simply the capacitive impedance that causes the angle of the impedance phasor to be 45:

$$\Z=90+160j-70j=90+90j\$$

and

$$\\theta_z=arctan(\frac{90}{90}) =45\$$

• "-70 is simply the capacitive impedance that causes the angle of the impedance phasor to be 45" can you show me the math?It will be very helpful.Thanks in advance and your for answer.I appreciate it a lot Commented Sep 10, 2023 at 14:13
• The math is given in the next two lines after that sentence... Commented Sep 11, 2023 at 12:30