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

Impedance and Generalized Ohm's Law

Generalized Ohm's Law

The Ohm's law $R=V/I$ in DC circuit composed of resistors only can be generalized to describe the relationship between the sinusoidal voltage and current associated with a component, a capacitor $C$, an inductor $L$, or a resistor $R$. The impedance of the component is defined as the ratio of the AC voltage and current associated with the component both represented as complex variables or phasors:

Z=\frac{{\bf V}(t)}{{\bf I}(t)}=\frac{V_p e^{j(\omega t+\phi...{\dot{V}}{\dot{I}}

The impedance $Z$ is in general complex and its magnitude is the ratio of the magnitudes of the voltage and the current, and its phase is the phase difference between the voltage and the current:

\vert Z\vert=\frac{\vert{\bf V}(t)\vert}{\vert{\bf I}(t)\ver...
...;\;\;\;\;\;\;\angle Z=\angle {\bf V}-\angle {\bf I}=\phi-\psi


The current $i(t)$ and voltage $v(t)$ associated with a component in an AC circuit are related by integration or differentiation in time domain. But if they are represented in phasor form $\dot{I}$ and $\dot{V}$ in frequency domain, they are related by algebraic equations. Phasor representation of a sinusoidal variable of angular frequency $\omega=2\pi f$ is the only Fourier coefficient in the frequency domain, the impedance can be considered as the frequency response function of the circuit component when the current through it and the voltage across it, both represented as phasors, are considered as the input and output of the component, respectively. As there is only one frequency in the signal, it does not need to be represented.


Impedance of Basic Components

The impedance of a specific component can be obtained according to the physics 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:

Impedance and Admittance

\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:

{\bf I}(t)=\frac{{\bf V}(t)}{Z}={\bf V}(t)Y

Sometimes it is more convenient in circuit analysis to use admittance instead of impedance.

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 2017-07-03