Inductors and Capacitors
Introduction: This article introduces inductance and capacitance within a circuit, namely through inductors and capacitors respectively.
The Essentials
A circuit element that we have really focused on is resistance, namely through resistors. There are other elements that behave similar to resistors, that being inductors and capacitors.
Let's first look at inductors. Inductors are described by inductance, seen as \( L \) in most equations. Inductance is measured in henrys (H). Inductor is represented as a coil of wire in many schematics. The inductor, due to the nature of the wire in a coil, stores energy in a surrounding magnetic field because of the flow of current. Figure 1. shows the schematic of an inductor.
Figure 1: Inductor Schematic Element
We are curious to know about the voltage and current through the inductor, which is described as the following
where
L = inductance in henrys
v = voltage in volts
i = current in amps
t = time in seconds
With this in mind, we describe the current as such:
where
L = inductance in henrys
\( t_0 \) = time in seconds
t = time in seconds
\( i(t_0) \) = initial current
For the current equation, we integrate over \( \tau \) simply as a filler variable because the current will be based on time \( t \), that is why we integrate from \( t_0 \) to \( t \). In most cases, \( t_0 = 0 \).
Other equations that could be helpful in analyzing the inductor are relating power and energy. Equation 3 describes power while equation 4 describes energy.
Now for the capacitor. Capacitors are described by capacitance, seen as \( C \) in equations. Capacitance is measured in farads (F). Capacitors are represented as two parallel, metal plates where there is no physical connection. The energy stored in the capacitor is possible by the electric field between the parallel plates. Figure 2. shows a schematic example of a capacitor.
Figure 2: Capacitor Schematic Element
We are curious about the current and voltage across the capacitor as well. The current described as the following:
where
i = current in amps
C = capacitance in farads
v = voltage in volts
t = time in seconds
Then for the voltage
where
i = current in amps
C = capacitance in farads
v = voltage in volts
t = time in seconds
We're also interested in the power and energy of a capacitor
These are a lot of equations to keep track of, but we'll do some practice to make sure we understand each of these new circuit elements.
Example
Let's first do examples of an inductor in circuits. We will do two examples, one involving the derivative of current and the integral of voltage. We will do the same thing for capacitance.
Figure 3. shows an example such a circuit.
Figure 3: Inductor Circuit Example
where
\( t < 0, \, i = 0 \, \text{A} \)
\( t > 0, \, i = 20te^{-5t} \, \text{A} \)
We are interested to know about the voltage across the inductor. Note that we are only curious to know about the voltage when \( t > 0 \) since current is 0 when \( t < 0 \), so we'll use the equation provided. Let's use equation 1 to solve this.
\[ v = L\frac{di}{dt} = 10\cdot10^{-3} \cdot \frac{d}{dt}(20te^{-5t}) = 0.2e^{-5t}(1-5t) \, \text{V}\]
What can help us is to see a graph of both the current and voltage over time. Figure 4. shows the current and voltage through the inductor. Blue shows the current and red shows the voltage.
Figure 4: Inductor Circuit Graphs (Example 1)
Notice how after a long time that as current decreases to 0, so does the voltage. This circuit simulates an instantaneous change in the circuit behavior, which we will explore more in the future and the relationship of an exponential decay. This is very common within circuitry.
Now let's another inductor problem, but let's say that we are given a voltage that is applied to an inductor. For this we are curious about the current through it. Let's assume the following:
\[ v(t) = 30te^{-10t} \, \text{V}\]
for
\( t > 0. \)
\( \text{Also assume} \, i = 0 \, \text{for} \, t \leq 0. \)
Now we are curious about the inductor current as a function of time. Let's say that the inductance of the inductor is 150 mH. To calculate this, we use equation 2. For this, we need to use integration by part. Reminder, we take the two separate terms (\( 30t \) and \( e^{-5t} \)) and we take derivatives and integrals of those to solve for a simpler term. We derive \( 30t \) and integrate \( e^{-5t} \) and we get the following.
\[ i(t) = \frac{1}{L}\int_{t_0}^t v \, d\tau + i(t_0) = \frac{1}{0.15} \int_0^t30\tau e^{-10\tau} \, d\tau + i(t_0) = \frac{1}{0.15}(\frac{3}{10}-3te^{-10t}-\frac{3}{10}e^{-10t})\, \text{A}\]
With the applied voltage and inductance, we get a current that stabilizes very quickly to 2 A. Again, a graph will help us understand both the voltage and current. Figure 5. contains that graph.
Figure 5: Inductor Circuit Graphs (Example 2)
Voltage is shown as blue while the current is red. Now as the voltage applied to the inductor decays exponentially, the current will settle to 2 A over a long period of time.
If we are interested in the power or energy within the inductor, we just apply equations 3 and 4 with our found calculations of v and i.
Now we have done examples with inductors, now let's do examples with capacitors, repeating a similar process but now with related capacitor values. Figure 6. has an example of a capacitor in series with a voltage source.
Figure 6: Capacitor Example
Let's say that the voltage applied across the capacitor is \( v(t) = 40e^{-20000t}\text{sin}(40000t) \) when \( t \geq 0 \). We'll assume that the voltage across the capacitor is 0 when \( t < 0 \). We are curious to know the current through the capacitor given the voltage across. We will use equation 5 to solve for the current.
\[ i(t) = C \frac{dv}{dt} = 75.2e^{-20000t}\text{cos}(40000t) - 37.6e^{-20000t}\text{sin}(40000t)\]
Let's look at a graph of current and voltage.
Figure 7: Capacitor Circuit Graph (Example 1)
This graph shows the voltage(blue) is shown as a exponentially decaying sinusoid along with the current(red). Note the time axis because these oscillations happen in very short time, within micro seconds. Both voltage and current dissipate to zero over a long time.
Let's do another example if capacitors but we are given the current applied to the capacitor. Let's say that the same capacitor from the previous problem is used and current is given by \( i(t) = 5\text{cos}(35000t) \, \text{A, for} \, t \geq 0 \). Assume that current is 0 A when \( t < 0 \). With the capacitor value from the previous problem (47 \( \mu \)F) and the current given, we can solve for the voltage \( v(t) \) across the capacitor.
\[ v(t) = \frac{1}{C}\int_0^t v(t)\, dt = \frac{1}{47 \cdot 10^{-6}}\int_0^t 5\text{cos}(35000t) \, dt = 3.04\text{sin}(35000t) \]
Graphing this, we see the following situation. The voltage is shown in red with a peak of about 3.04 volts and current is shown in blue with a peak of 5 amps. Figure 8. has an image of this graph.
Figure 8: Capacitor Circuit Graph (Example 2)
All being said and done, we can find the current and voltage through either a capacitor or inductor with the help of derivatives and integrals.
Practice
- Given a current \( i(t) = 200t \, \text{A} \) is applied to a 28 mH inductor in series, solve for the voltage across the inductor.
- Given a current \( i(t) = 50te^{-5t} \, \text{A} \) is applied to a 15 mH inductor in series, solve for the voltage across the inductor. Also find the power stored within the inductor
- Given a voltage \( v(t) = 60t\text{cos}(2500t) \, \text{A} \) is applied to a 20 mH inductor in series, solve for the current through the inductor. Assume the current at time 0 is 0 A.
- Given a the voltage \( v(t) = 3t \, \text{V} \) is applied to a 10 mF capacitor in series, solve for the current through the capacitor.
- Given a the voltage \( v(t) = 10t\text{cos}(40000t) \) is applied to a 15 \( \mu \)F capacitor in series, solve for the current through the capacitor. Also find the power stored within the capacitor.
- Given a current \( i(t) = 30te^{-10t} \) is applied to a 250 \( \mu \)F capacitor in series, solve for the voltage across the capacitor. Assume the voltage at time 0 is 300 V.
Solutions:
Problem 1 Solution.) \( v(t) = 5.6 \, \text{V} \). Using equation 1 we get the following.
\[ v(t) = 28\cdot10^{-3} \frac{d}{dt}(200t) = 5.6 \, \text{V}\]
Problem 2 Solution.) \( v(t) = 0.75e^{-5t}(1-5t) \, \text{V} \) and \( p(t) = 36te^{-10t}(1-5t) \). Using equation 1,
\[ v(t) = 15\cdot10^{-3}\frac{d}{dt}(50te^{-5t}) = 15\cdot10^{-3}(50e^{-5t} - 250te^{-5t}) = 0.75e^{-5t}(1-5t) \, \text{V}\]
To find the power within the inductor, we use equation 3. It can also be rewritten as \( p = v \cdot i \).
\[ p = v \cdot i = 0.75e^{-5t}(1-5t) \cdot 50te^{-5t} = 36te^{-10t}(1-5t) \, \text{W} \]
Problem 3 Solution.) \( i(t) = \frac{6}{5}(t\text{sin}(2500t) - \frac{\text{cos}(2500t)}{2500} + \frac{1}{2500}) \). We are given the voltage across the inductor and it's inductance. We use equation 2 to solve for the current through it. We need to use integration by parts to solve for the integral of \( v(t) \).
\[ i(t) = \frac{1}{20\cdot 10^{-3}} \int_0^t60t\text{cos}(2500t)\, dt = 3000(\frac{t\text{sin}(2500t)}{2500} - \frac{\text{cos}(2500t)}{2500^2} + \frac{1}{2500}) = \frac{3000}{2500}(t\text{sin}(2500t) - \frac{\text{cos}(2500t)}{2500} + \frac{1}{2500})\]
Problem 4 Solution.) \( i(t) = 30 \, \text{mA} \). Use equation 5 to solve for current.
\[ i = 10\cdot10^{-3} \frac{d}{dt}(3t) = 0.03 \, \text{A}\]
Problem 5 Solution.) \( i(t) = 15\cdot10^{-5}(\text{cos}(40000t) - 40000t\text{sin}(40000t)) \) and \( p(t) = 15\cdot10^{-4}t\text{cos}(40000t)(\text{cos}(40000t) - 40000t\text{sin}(40000t)) \). Given the voltage, we can solve for the current through the capacitor by using equation 5.
\[ i = 15 \cdot 10^{-6} \frac{d}{dt}(10t\text{cos}(40000t)) = 15\cdot10^{-6}(10\text{cos}(40000t)-40000t\text{sin}(40000t))\]
With the current equation, now we can solve for the power stored in the capacitor using equation 7.
\[ p(t) = 15\cdot10^{-6}(10\text{cos}(40000t)-40000t\text{sin}(40000t)) \cdot 10t\text{cos}(40000t) = 15\cdot10^{-4}t\text{cos}(40000t)(\text{cos}(40000t) - 40000t\text{sin}(40000t))\]
Problem 6 Solution.) \( v(t) = 120000(\frac{1}{100}-\frac{te^{-10t}}{10}-\frac{e^{-10t}}{{100}}) + 300 \, \text{V} \). We use equation 6 to solve for the voltage equation given the current.
\[ v(t) = \frac{1}{250\cdot10^{-6}}\int_0^t30te^{-10t}\, dt = \frac{30}{250\cdot10^{-6}}(-\frac{te^{-10t}}{10}-\frac{e^{-10t}}{100}) \Big|_0^t\]
Evaluate this integral and you should most of the equation above. Remember that we add on our initial conditions, which in this case, \( v(0) = 300 \, \text{V} \). Just add that to the result of the integral to get the final answer.