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\$.
Best Answer
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.)