AC Circuit Analysis I (Steady State)

While DC circuit analysis is carried out by solving algebraic equations, the analysis of AC circuits composed of capacitors, inductors as well as resistors will require solving differential equations (DEs). The solution of a DE represent the response (or output) of the circuit to both the external input and the initial state, and is composed of two parts:

• Homogeneous solution representing the natural response caused by the initial condition,
• Particular solutions representing the forced response caused by the external input.

If only the steady state solutions are of interest, the phasor method can be used to solve the problem algebraically without solving the DEs. Specifically, all sinusoidal variables are represented as phasors in terms of their amplitudes and phases, and all components in the circuit (inductors and capacitors as well as resistors) are represented by their impedances, so that all the laws (Ohm's law, KCL and KVL, current and voltage dividers, parallel and series combinations of components) and methods (loop current and node voltage methods, Thevenin's and Norton's theorems, etc.) discussed for DC circuit apply.

Example 1: Given an RL circuit consisted of a resistor and an inductor in series connected to an AC voltage source $v(t)=\cos(\omega t)=Re[e^{j\omega t}]$, find the current $i(t)$.

Method 1: The current $i(t)$ can be found by solving the governing equation (a DE) below that describes the circuit:

$$v_L(t)+v_R(t)=L\frac{d}{dt}i(t)+R\;i(t)=v(t)=\cos(\omega t) =\frac{1}{2}(e^{j\omega t}+e^{-j\omega t})$$

We want to find its particular or steady state solution. (The homogeneous or transient solution will be discussed later.) Due to Euler's formula, Since the DE describes a linear system, the superposition principle applies $O(ax+by)=aO(x)+bO(y)$. If we find the solutions for

$$L\frac{d}{dt}i_1(t)+Ri_1(t)=v_1(t)=e^{j\omega t},\;\;\;\;\; ... ...}\;\;\;\;\; L\frac{d}{dt}i_2(t)+Ri_2(t)=v_2(t)=e^{-j\omega t}$$

we can then find the solution for the original input as $i(t)=[i_1(t)+i_2(t)]/2$. To solve the DE with input $v_1(t)=e^{j\omega t}$, we assume

$$i_1(t)=A\;e^{j\omega t},\;\;\;\;\;\frac{d}{dt}i_1(t)=j\omega A\;e^{j\omega t}$$

and substitute them back into the DE to get

$$j\omega L A e^{j\omega t}+RA e^{j\omega t} =(R+j\omega L)A\; e^{j\omega t}=e^{j\omega t}$$

i.e.

$$A=\frac{1}{R+j\omega L}=\vert A\vert e^{-j\phi},\;\;\;\;\mbox... ...2 L^2}},\;\;\;\;\; \phi=\angle A=tan^{-1}(\frac{\omega L}{R})$$

and the solution is:

$$i_1(t)=A\;e^{j\omega t}=\frac{1}{\sqrt{R^2+\omega^2 L^2}} e^{j(\omega t-\phi)}$$

Similarly, when the input is $v_2(t)=e^{-j\omega t}$ we assume

$$i_2(t)=A\;e^{-j\omega t},\;\;\;\;\;\frac{d}{dt}i_2(t)=-j\omega A\;e^{-j\omega t}$$

and substitute them back into the DE to get

$$-j\omega L A e^{-j\omega t}+RA e^{-j\omega t} =(R-j\omega L)A\; e^{-j\omega t}=e^{-j\omega t}$$

i.e.

$$A=\frac{1}{R-j\omega L}=\vert A\vert e^{j\phi}$$

with $\vert A\vert$ and $\phi$ the same as defined above. The solution is:

$$i_2(t)=A\;e^{-j\omega t}=\frac{1}{\sqrt{R^2+\omega^2 L^2}} e^{-j(\omega t-\phi)}$$

The solution (the response or output) for the DE with input $v(t)=[v_1(t)+v_2(t)]/2$ is

$$i(t)=\frac{i_1(t)+i_2(t)}{2}=\vert A\vert\;\frac{e^{j(\omega ... ...i)}}{2}=\frac{1}{\sqrt{R^2+\omega^2 L^2}} \cos(\omega t-\phi)$$

Method 2: Since only the steady state solution is of interest, the phasor method can be used. Specifically, here the current $i(t)$ can be much more easily found by generalized Ohm's law. The complex representation of the input voltage is

$${\bf V}=e^{j\omega t}$$

The overall complex impedance of the two elements in series is:

$$Z=Z_R+Z_L=R+j\omega L=\vert Z\vert e^{j\angle Z}=\sqrt{R^2+\omega^2L^2}e^{j \phi}$$

where $\phi=tan^{-1} (\omega L/R)$. By generalized Ohm's law, the complex representation of the current can be found to be:

$${\bf I}=\frac{{\bf V}}{Z} =\frac{e^{j\omega t}}{\sqrt{R^2+\o... ...\phi}} =\frac{1}{\sqrt{R^2+\omega^2L^2}} e^{j(\omega t-\phi)}$$

and the real current is

$$i(t)=Re[{\bf I}]=Re[\frac{1}{\sqrt{R^2+\omega^2 L^2}}e^{j(\om... ...\phi)}] =\frac{1}{\sqrt{R^2+\omega^2 L^2}}\cos(\omega t-\phi)$$

Also recall that in general, the frequency response function (FRF) of the system is defined as the ratio of the output to the input of the system, both represented as complex exponentials. In this specific case, we have

$$H=\frac{{\bf I}}{{\bf V}}=\frac{1}{Z}$$

with

$$\vert H\vert=\frac{1}{\sqrt{R^2+\omega^2L^2}},\;\;\;\;\;\angle H=-\phi$$

Therefore the steady state output can be found to be:

$$i(t)=\vert H\vert \cos(\omega t+\angle H) =\frac{1}{\sqrt{R^2+\omega^2 L^2}}\cos(\omega t-\phi)$$

The second method, much easier than the first one, is actually a short cut representation of the first DE method. This is the justification of the complex variable or phasor method for analyzing AC circuits. However, note that the phasor method can only find the steady state solution. The homogeneous differential equation will have to be solved to obtain the transient solution.

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.

• Express $i(t)$ in phasor: $\dot{I}=12\angle{90^\circ}$
• Find impedance for each element ($\omega=1000$):
  
  $$Z_R=R=18,\;\;\;\;Z_C=1/j\omega C=12\angle{-90^\circ}=0-j12,\;\;\;\; Z_L=j\omega L=30\angle{90^\circ}=0+j30$$
  
  

• Find overall impedance:
  
  $$Z_{total}=Z_R+Z_C+Z_L=18+j(-12+30)=18+j18=18\sqrt{2}\angle{45^\circ}$$
  
  

• Find voltage across all three elements:
  
  $$\dot{V}_{total}=\dot{I}Z_{total}=(12\angle{90^\circ})\;(18\sqrt{2}\angle{45^\circ})=216\sqrt{2}\angle{135^\circ}$$
  
  
  
  
  $$v(t)=(216\sqrt{2})\sqrt{2}cos(1000t+135^\circ)=432\;cos(1000t+135^\circ)$$
  
  

Alternatively, we can find voltage across each element:

$$V_R=IZ_R=216\angle{90^\circ},\;\;\;\; V_C=IZ_C=144\angle{0^\circ},\;\;\;\; V_L=IZ_L=360\angle{180^\circ}$$

and the total voltage is

$$\dot{V}_{total}=V_R+V_C+V_L=216\angle{90^\circ}+216\angle{180^\circ} =216\sqrt{2}\angle{135^\circ}$$

Example 3: The circuit below represents a load consisting of $C$, $R$ and $L$ supplied by a generator over a transmission line. 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

• $Y_C=j\omega C=j\;5000\times 10^{-5}=j0.05=0.05\angle{90^\circ}$,

  $Z_C=1/Y_C=1/j0.05=-j20=20\angle{-90^\circ}$

• $Z_L=j\omega L=j5000\times 4\times 10^{-3}=0+j20=20\angle{90^\circ}$,

  $Y_L=1/Z_L=-j0.05=0.05\angle{-90^\circ}$

• $Z_R=R=20+j0=20\angle{0^\circ}$

• $Z_{RL}=Z_R+Z_L=20+j20=20\sqrt{2}\angle{45^\circ}$, $Y_{RL}=1/Z_{RL}=1/20\sqrt{2}\angle{45^\circ} =0.025\sqrt{2}\angle{-45^\circ}=0.025-j0.025$

• $Y_{load}=Y_C+Y_{RL}=j0.05+0.025-j0.025=0.025+j0.025 =0.025\sqrt{2}\angle{45^\circ}$

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:

• $\dot{I}_{load}=Y_{load}\dot{V}=(20\angle{45^\circ})\; (0.025\sqrt{2}\angle{45^\circ})=0.5\sqrt{2}\angle{90^\circ}$,

  $i_{load}(t)=Re[\dot{I}\sqrt{2}e^{j5000t}]=cos(5000t+90^\circ)$

• $\dot{I}_C=\dot{V}Y_C=(0.05\angle{90^\circ})(20\angle{45^\circ}) =1\angle{135^\circ}=-0.5\sqrt{2}+j0.5\sqrt{2}$,

  $i_C(t)=\sqrt{2}cos(5000t+135^\circ)$

• $\dot{I}_{RL}=Y_{RL}\dot{V}=(0.025\sqrt{2}\angle{-45^\circ})(20\angle{45^\circ}) =0.5\sqrt{2}\angle{0^\circ}=0.5\sqrt{2}$,

  $i_{RL}(t)=cos(5000t)$

• $\dot{V}_R=Z_R\dot{I}_{RL}=(20\angle{0^\circ})(0.5\sqrt{2}\angle{0^\circ}) =10\sqrt{2}\angle{0^\circ}=10\sqrt{2}+j0$, $i_R(t)=20\;cos(5000t)$
• $\dot{V}_L=Z_L\dot{I}_{RL}=(20\angle{90^\circ})(0.5\sqrt{2}\angle{0^\circ}) =10\sqrt{2}\angle{90^\circ}=0+j10\sqrt{2}$, $v_L(t)=20\;cos(5000t+90^\circ)$.

Verify:

• $\dot{V}_R+\dot{V}_L=10\sqrt{2}+j10\sqrt{2}=20\angle{45^\circ} =\dot{V}$
• $\dot{I}_C+\dot{I}_{RL}=0.5\sqrt{2}-0.5\sqrt{2}+j0.5\sqrt{2} =0.5\sqrt{2}\angle{90^\circ}=\dot{I}_{load}$