4.5 First-order linear equations  (Page 5/10)

Electrical circuits

A source of electromotive force (e.g., a battery or generator) produces a flow of current in a closed circuit, and this current produces a voltage drop across each resistor, inductor, and capacitor in the circuit. Kirchhoff’s Loop Rule states that the sum of the voltage drops across resistors, inductors, and capacitors is equal to the total electromotive force in a closed circuit. We have the following three results:

1. The voltage drop across a resistor is given by
   
   $E_R=Ri,$
   
   where $R$ is a constant of proportionality called the resistance, and $i$ is the current.
2. The voltage drop across an inductor is given by
   
   $E_L=L{i}^{\prime },$
   
   where $L$ is a constant of proportionality called the inductance , and $i$ again denotes the current.
3. The voltage drop across a capacitor is given by
   
   $E_C=\frac{1}{C}q,$

where $C$ is a constant of proportionality called the capacitance , and $q$ is the instantaneous charge on the capacitor. The relationship between $i$ and $q$ is $i=q^{\prime }.$

We use units of volts $\left(\text{V}\right)$ to measure voltage $E,$ amperes $\left(\text{A}\right)$ to measure current $i,$ coulombs $\left(\text{C}\right)$ to measure charge $q,$ ohms $\left(\text{Ω}\right)$ to measure resistance $R,$ henrys $\left(\text{H}\right)$ to measure inductance $L,$ and farads $\left(\text{F}\right)$ to measure capacitance $C.$ Consider the circuit in [link] .

Applying Kirchhoff’s Loop Rule to this circuit, we let $E$ denote the electromotive force supplied by the voltage generator. Then

Substituting the expressions for $E_L,E_R,$ and $E_C$ into this equation, we obtain

If there is no capacitor in the circuit, then the equation becomes

This is a first-order differential equation in $i.$ The circuit is referred to as an $LR$ circuit.

Next, suppose there is no inductor in the circuit, but there is a capacitor and a resistor, so $L=0,R\ne 0,$ and $C\ne 0.$ Then [link] can be rewritten as

which is a first-order linear differential equation. This is referred to as an RC circuit . In either case, we can set up and solve an initial-value problem.

Finding current in an RL Electric circuit

A circuit has in series an electromotive force given by $E=50\text{sin}20t\text{V},$ a resistor of $5\text{Ω},$ and an inductor of $0.4\text{H}\text{.}$ If the initial current is $0,$ find the current at time $t>0.$

We have a resistor and an inductor in the circuit, so we use [link] . The voltage drop across the resistor is given by $E_R=Ri=5i.$ The voltage drop across the inductor is given by $E_L=L{i}^{\prime }=0.4i^{\prime }.$ The electromotive force becomes the right-hand side of [link] . Therefore [link] becomes

$0.4i^{\prime }+5i=50\text{sin}20t.$

Dividing both sides by $0.4$ gives the equation

$i^{\prime }+12.5i=125\text{sin}20t.$

Since the initial current is 0, this result gives an initial condition of $i\left(0\right)=0.$ We can solve this initial-value problem using the five-step strategy for solving first-order differential equations.

Step 1. Rewrite the differential equation as $i^{\prime }+12.5i=125\text{sin}20t.$ This gives $p\left(t\right)=12.5$ and $q\left(t\right)=125\text{sin}20t.$

Step 2. The integrating factor is $\mu \left(t\right)=e^{{\displaystyle \int 12.5dt}}=e^{12.5t}.$

Step 3. Multiply the differential equation by $\mu \left(t\right)\text{:}$

$\begin{array}{ccc}\hfill e^{12.5t}{i}^{\prime }+12.5e^{12.5t}i& =\hfill & 125e^{12.5t}\text{sin}20t\hfill \\ \hfill \frac{d}{dt}\left[i{e}^{12.5t}\right]& =\hfill & 125e^{12.5t}\text{sin}20t.\hfill \end{array}$

Step 4. Integrate both sides:

$\begin{array}{ccc}\hfill {\displaystyle \int \frac{d}{dt}\left[i{e}^{12.5t}\right]}dt& =\hfill & {\displaystyle \int 125e^{12.5t}\text{sin}20tdt}\hfill \\ \hfill i{e}^{12.5t}& =\hfill & \left(\frac{250\text{sin}20t-400\text{cos}20t}{89}\right)e^{12.5t}+C\hfill \\ \hfill i\left(t\right)& =\hfill & \frac{250\text{sin}20t-400\text{cos}20t}{89}+C{e}^{\mathrm{-12.5}t}.\hfill \end{array}$

Step 5. Solve for $C$ using the initial condition $v\left(0\right)=2\text{:}$

$\begin{array}{ccc}\hfill i\left(t\right)& =\hfill & \frac{250\text{sin}20t-400\text{cos}20t}{89}+C{e}^{\mathrm{-12.5}t}\hfill \\ \hfill i\left(0\right)& =\hfill & \frac{250\text{sin}20\left(0\right)-400\text{cos}20\left(0\right)}{89}+C{e}^{\mathrm{-12.5}\left(0\right)}\hfill \\ \hfill 0& =\hfill & -\frac{400}{89}+C\hfill \\ \hfill C& =\hfill & \frac{400}{89}.\hfill \end{array}$

Therefore the solution to the initial-value problem is $i\left(t\right)=\frac{250\text{sin}20t-400\text{cos}20t+400e^{\mathrm{-12.5}t}}{89}=\frac{250\text{sin}20t-400\text{cos}20t}{89}+\frac{400e^{\mathrm{-12.5}t}}{89}.$

The first term can be rewritten as a single cosine function. First, multiply and divide by $\sqrt{{250}^2+{400}^2}=50\sqrt{89}\text{:}$

$\begin{array}{cc}\hfill \frac{250\text{sin}20t-400\text{cos}20t}{89}& =\frac{50\sqrt{89}}{89}\left(\frac{250\text{sin}20t-400\text{cos}20t}{50\sqrt{89}}\right)\hfill \\ & =-\frac{50\sqrt{89}}{89}\left(\frac{8\text{cos}20t}{\sqrt{89}}-\frac{5\text{sin}20t}{\sqrt{89}}\right).\hfill \end{array}$

Next, define $\phi$ to be an acute angle such that $\text{cos}\phi =\frac{8}{\sqrt{89}}.$ Then $\text{sin}\phi =\frac{5}{\sqrt{89}}$ and

$\begin{array}{cc}\hfill -\frac{50\sqrt{89}}{89}\left(\frac{8\text{cos}20t}{\sqrt{89}}-\frac{5\text{sin}20t}{\sqrt{89}}\right)& =-\frac{50\sqrt{89}}{89}\left(\text{cos}\phi \text{cos}20t-\text{sin}\phi \text{sin}20t\right)\hfill \\ & =-\frac{50\sqrt{89}}{89}\text{cos}\left(20t+\phi \right).\hfill \end{array}$

Therefore the solution can be written as

$i\left(t\right)=-\frac{50\sqrt{89}}{89}\text{cos}\left(20t+\phi \right)+\frac{400e^{\mathrm{-12.5}t}}{89}.$

The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is given by $\phi ,$ and the amplitude of the steady-state current is given by $\frac{50\sqrt{89}}{89}.$ The graph of this solution appears in [link] :

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