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Next: First Order Systems Up: Chapter 3: AC Circuit Previous: Phasor Representation of Sinusoidal

Impedance and Generalized Ohm's Law

Impedance of Basic Components

The relationship between the sinusoidal current $i(t)$ through and the sinusoidal voltage $v(t)$ across a capacitor or an inductor in an AC circuit is described by a differential equation in time domain. However, if we treat such a sinusoidal variable as the projection of a complex exponential, a vector rotating counter-clock wise, onto the real axis:

v(t)=V_p\cos(\omega t+\phi)=Re[ V_p e^{j(\omega t+\phi)} ],
i(t)=I_p\cos(\omega t+\psi)=Re[ I_p e^{j(\omega t+\psi)} ]

then the relationship between the voltage and current can be described by an algebraic equation. Specifically, we define the ratio of the complex exponential forms of the voltage and the current associated with a resistor $R$, a capacitor $C$, or an inductor $L$ as the impedance of the component:

One way to remember the phase between the voltage $E$ and current $I$ associated with capacitor $C$ and inductor $L$ is ``ELI the ICE man''. Also, consider two extreme cases:

In any of the cases above, the impedance is defined as

=\frac{\mbox{Phasor of voltage}}{\mbox{Phasor of current}}

This is the generalized version of the Ohm's law for AC circuits.


\mbox{Impedance} &=& \mbox{Resist...
... \mbox{Admittance}&=&1/\mbox{Impedance}\\

Impedance $Z$ and admittance $Y=1/Z$ are both complex variables. The real parts $Re[Z]=R$ and $Re[Y]=G$ are always positive, but the imaginary parts $Im[Z]=X$ and $Im[Y]=B$ can be either positive or negative. Therefore $Z$ and $Y$ can only be in the 1st or the 4th quadrants of the complex plane.

In particular, the admittances of the three types of elements R, L and C are

Y_L=\frac{1}{Z_L}=\frac{1}{j\omega L...
... L},\;\;\;\;
Y_C=\frac{1}{Z_C}=\frac{1}{1/j\omega C}=j\omega C \end{displaymath}

Ohm's law can also be expressed in terms of admittance as well as impedance. Sometimes it is more convenient in circuit analysis to use admittance instead of impedance.

Here is a review of complex arithmetic.

Example 1:

Solve the circuit below. The voltage from the generator is $v(t)=28.3\;\cos(5000t+45^\circ)\;V$.


First find the impedances and admittances of the components and the two branches. As $\omega=5000$, we get

Next express voltage $v(t)=28.3\;\cos(5000t+45^\circ)\;V$ in phasor form $\dot{V}=20\angle{45^\circ}$, and find currents in phasor form: Verify:


Example 2:

A current $i(t)=17\;\cos(1000t+90^\circ)=12\sqrt{2}\cos(1000t+90^\circ)$ flows through a circuit composed of a resistor $R=18\Omega$, and capacitor $C=83.3\mu F=83.3\times 10^{-6}F$ and an inductor $L=30 mH=30\times
10^{-3}H$ connected in series. Find the resulting voltage across all three elements.

Alternatively, we can find voltage across each element:


In time domain, we have


Adding $\dot{V}_R$, $\dot{V}_C$, and $\dot{V}_L$, we get the total voltage which is the same as what we got above:


Example 3:

In the circuit below, $v_S(t)=A\cos(\omega t)$ ($\omega=10^6$) with some unknown peak value $A$, $R_1=R_2=R_3=R=10\Omega$, and $L=10 \mu H$. The RMS value of $v_2$ across $R_2$ is 10 V. It is also known that $v_s(t)$ and $i_s(t)$ are in phase.




Note that $\dot{V}_C$ is behind $\dot{V}_2=\dot{I}_2R_2$ by $\pi/2$, and $\dot{V}_L$ is ahead of $\dot{V}_3=\dot{I}_3R_3$ by $\pi/2$ (``ELI the ICE man''). Also, as $v_s(t)$ and $i_s(t)$ are in phase, the impedance of the parallel combination of the RL and RC branches shown below must real introducing no phase shift:

Z_{RL}(\omega)\vert\vert Z_{RC}(\omega)

i.e., $X_C=-X_L$ or $1/j\omega C=-j\omega L=j 10^7$. We therefore get: $C=10^{-7}=0.1\mu F$, $Z_{RL}=10+j 10$, $Z_{RC}=10-j 10$, $\vert Z_{RL}\vert=\vert Z_{RC}\vert=\sqrt{10^2+10^2}=10\sqrt{2}$, and

Z_{RL}(\omega)\vert\vert Z_{RC}(\omega)=\frac{(10+j10)(10-j10)}{10+j10+10-j10}=10

As $\vert R_2\vert=\vert Z_C\vert=10 \Omega$, $\vert\dot{V}_C\vert=\vert\dot{V}_R\vert=10V$. But as they are $\pi/2$ apart in phase, we have $\vert\dot{V}_{cd}=\vert\dot{V}_{RC}\vert=\vert\dot{V}_{RL}\vert=10\sqrt{2}$, and from the vector diagram $\vert\dot{V}_{ab}\vert=10\sqrt{2}$. We also get the currents through RC and RL branches are:

\vert\dot{I}_{RC}\vert=\frac{\vert\dot{V}_{RC}\vert}{\vert Z...
..._{RL}\vert}{\vert Z_{RL}\vert}=\frac{10\sqrt{2}}{10\sqrt{2}}=1

But their phase difference is $\pi/2$, we have


The voltage across $R_1$ is $V_1=RI_s=10\sqrt{2}$, and


The peak value is therefore $A=\sqrt{2} V_s=40 V$

next up previous
Next: First Order Systems Up: Chapter 3: AC Circuit Previous: Phasor Representation of Sinusoidal
Ruye Wang 2018-04-02