[[The Bare Essentials of Electrical Engineering.pdf]] # Capacitors - Capacitors are passive elements that store electric field energy - Parallel-plate image consists of two parallel metal sheets w/ insulator between them. When a voltage difference is applied to the plates, negative charges accumulate at one plate while positive charges accumulate on the opposite side - Opposite charges on the parallel plates create an electric field in between them where electric energy is stored. Amount of charge $q\propto$ applied voltage $V$, constant of proportionality $C=\frac{q}{V}$ - Units of capacitance given in farad $F$ where $1F=\frac{1C}{1V}$ - $C$ depends solely on the geometry of the structure and its material properties - Can increase capacitance of parallel plate capacitor by decreasing distance between plates/increasing surface area/changing material: $C=\frac{\epsilon A}{d}$ - An ideal capacitor holds uniform charge distribution indefinitely and has an electric field $\mathcal{E}$ in straight lines (see [[The Bare Essentials of Electrical Engineering.pdf#page=160|figure 5.2]]) ## Finding the Current Through A Capacitor - From the current definition we know that $i_{C}(t)=\frac{dq}{dt}$, taking the derivative of both sides and assuming constant capacitance yields $\frac{dq}{dt}=CB \frac{dv_{C}(t)}{dt}\implies \boxed{i_{c}(t)=C \frac{dv_{C}(t)}{dt}}$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=163|Example 5.2.2]] - Voltage of a capacitor must be continuous and cannot change abruptly because a sudden jump in it voltage would result in infinite current $\implies \forall t,v_{C}(t^-)=v_{C}(t^+)$ - Just because the voltage cannot change abruptly does not mean that the current will behave in the same manner - Capacitor behaves as an open circuit w/ DC power because $i_{C}(t)=C \frac{dv_{C}(t)}{dt}=C \cdot0=0A$ - Can still have voltage drop across it - even w/ zero current, voltage across the capacitor is $v_{c}=\frac{R_{2}}{R_{1}+R_{2}}V_{s},$ see [[The Bare Essentials of Electrical Engineering.pdf#page=166|Example 5.2.3]], [[The Bare Essentials of Electrical Engineering.pdf#page=167|5.2.4]] ## Finding the Voltage Across a Capacitor > [!derivation] > $dv_{C}(t)=\frac{1}{C}i_{C}(t)dt$, integrating both sides and changing integration values we see: $\int _{-\infty }^{v_{C}(t)}dv_{C}=\frac{1}{C}\int _{-\infty} ^{t}i_{C}(x)dx\to v_{C}(t)=\frac{1}{C} \int _{-\infty }^ti_{C}(x)dx$ > Assume everything is constant until some time $t_{0}$, when a switch is flipped or a source is turned on/off: > $v_{C}(t)=\underbrace{ \frac{1}{C} \int _{-\infty} ^{t_{0}}i_{C}(x)dx }_{ A }+\underbrace{ \frac{1}{C}\int _{t_{0}} ^{t}i_{C}(x)dt }_{ B }$ > Part $A$ represents voltage across the capacitor at the time $t_{0}$ and we can use $v_{C}(t_{0})$ to represent the voltage, often called the initial condition -> $\boxed{ v_{C}(t)=\frac{1}{C} \int _{t_{0}} ^t i_{C}(x)dx+v_{C}(t_{0}) }^{ }$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=161|Example 5.2.5, Example 5.2.6]] ## Energy Stored in a Capacitor > [!derivation] > Substituting capacitance equation into power equation $p_{C}(t)=v_{C}(t)i_{C}(t)=v_{C}(t)C \frac{dv_{C}(t)}{dt}$ > To find energy stored in the capacitor, we integrate the power over which the capacitor is charging: > $w_{C}(t)=\int _{{-\infty}}^tp_{C}(x)dx=C\int _{{-\infty}}^t v_{C}(x) \frac{dv_{C}(x)}{dx}dx=\frac{C}{2}v_{C}^2(x)\Biggr|^t _{-\infty}=\frac{1}{2}C(v_{C}^2(t)-v_{C}^2(-\infty))$ > Assuming the capacitor is not charged at $t=-\infty,(v_{C}(t=\infty)=0),$ we can write $\boxed{ w_{C}(t)=\frac{1}{2}Cv_{C}^2(t) }^{ }$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=171|Example 5.2.7]] ## Capacitor Connections > [!derivation] > For capacitors in series, take KVL around the circuit in [[The Bare Essentials of Electrical Engineering.pdf#page=174|Figure 5.20]] then substitute $v_{C}(t)=\frac{1}{C}\int_{t_{0}}^ti_{C}(x)dx+v_{C}(t_{0}):$ > $v_{s}(t)=\frac{1}{C_{1}}\int _{t_{0}}^t i_{C}(x)dx+v_{C_{1}}(t_{0})+\frac{1}{C_{2}} \int _{t_{0}} ^t i_{C}(x)dx+v_{C_{2}}(t_{0})+\dots+\frac{1}{C_{N}}\int _{t_{0}}^t i_{C}(x)dx+v_{C_{N}}(t_{0})$ > $=\left( \frac{1}{C_{1}}+\frac{1}{C_{2}}+\dots+\frac{1}{C_{N}} \right)\int _{t_{0}}^t i_{C}(x)dx+v_{C_{1}}(t_{0})+v_{C_{2}}(t_{0})+\dots+v_{C_{N}}(t_{0})$ > $=\frac{1}{C_{eq}}\int _{t_{0}}^ti_{C}(x)dx+v_{C_{1}}(t_{0})+v_{C_{2}}(t_{0})+\dots+v_{C_{N}}(t_{0})\text{ where } \boxed{ \frac{1}{C_{eq}}=\sum^N _{{i=1}} \frac{1}{C_{i}} }^{ }$ > [!derivation] >For capacitors in parallel, take KCL around the circuit in [[The Bare Essentials of Electrical Engineering.pdf#page=175|Figure 5.2.1]] then substitute $i_{C}(t)=C \frac{dv_{C}(t)}{dt}$ -> since the voltage is the same across all capacitors, we have > $i_{C}=C_{1} \frac{dv(t)}{dt}+C_{2} \frac{dv(t)}{dt}+\dots+C_{N} \frac{dv(t)}{dt}=\frac{dv(t)}{dt}(C_{1}+C_{2}+\dots+C_{N})$ > $=C_{eq} \frac{dv(t)}{dt}\text{ where } \boxed{ C_{eq}=\sum^N _{i=1}C_{n} }^{ }$ - See [[The Bare Essentials of Electrical Engineering.pdf#page=176|Example 5.2.9]], [[The Bare Essentials of Electrical Engineering.pdf#page=177|Example 5.2.10]] # Inductors - Inductors are passive elements that store magnetic energy. Inductance is measure in Henries (H) - vary widely in shapes and sizes depending on application ## Self Inductance - Consider a loop with a current $I$, creating a magnetic field $\vec{B}.$ Magnetic field penetrates the surface area $S$ of the loop resulting in a magnetic flux $\Phi$ through the loop defined by $\Phi=\int _S \vec{B}\cdot d\vec{S}$. Magnetic field is measured in Teslas ($T$) and $dS$ in $m^2$ - Self-inductance, or inductance of the current loop is defined as $L=\frac{\Phi}{I}$ and if we had $N$ turns of the loop, the inductance would be given by $L=\frac{N\Phi}{I}=\frac{\lambda}{I}$ - In the inductance defining equation, $I$ is in the denominator and associated flux $\Phi \propto \vec{B}$ and $\vec{B}\propto I\therefore \frac{\Phi}{I}=C$ so inductors don't depend on current/voltage, only geometry and material - Considering a solenoid with $N$ turns, length $l$, surface area $S$, and current $I$, we find that $L=N\mu_{0}nS=N\mu_{0} \frac{N}{l}S=\mu_{0} \frac{N^2}{l}S$ once again proving that the inductance depends only on the geometry of the material, and number of turns can increase the inductance dramatically ([[The Bare Essentials of Electrical Engineering.pdf#page=179|page 179]]) ## The Induced Voltage of Electromotive Force (emf) - Faraday's Law: a current is inducted in a closed loop if we have a varying magnetic field through the loop even in the absence of any explicit circuit source ([[The Bare Essentials of Electrical Engineering.pdf#page=180|page 180]]) - Existence of this current is explained by electromotive force, essentially an induced voltage around the loop generating the current. If loop is open, there is no current but there will still be induced emf $\oint_{c}\vec{E}\cdot d \vec{\ell}=-\frac{d}{dt}\int _{S}\vec{B}\cdot d\vec{S}\implies V_{emf}=-\frac{d}{dt}\Phi$ - Equation above shows that change in magnetic flux results in an induced voltage. Negative sign indicates that the induced voltage generates a current whose magnetic field *opposes the charge caused by the external magnetic field* - See [[The Bare Essentials of Electrical Engineering.pdf#page=181|Example 5.3.1]] - Remember right hand rule for determining current direction - Can consider the schematic as what's going on "under the hood" of an inductor - Induced voltage $v(t)=\frac{d\Phi}{dt}$ where $\Phi=Li(t)$ so the voltage across an inductor is given by $\boxed{ v_{L}(t)=L \frac{di_{L}(t)}{dt} }^{ },$ known as the basic current-voltage characteristic of the inductor - Indicated passive sign convention is consistent with Lenz's law. Consider $i(t)$ being imposed by an external source and induced voltage $v(t)$ has correct polarity. Considering case where $i(t)$ increases with time, then $v(t)$ with the indicated sign convention should be positive, since it would tend to induce an opposite current to oppose the imposed change. ([[The Bare Essentials of Electrical Engineering.pdf#page=183|page 175]]) - In the presence of an increasing $i(t),$ the inductor would build up a potential barrier by storing magnetic field energy. The externally imposing current would need to overcome this potential barrier to continue flowing -> the higher the rate of change of the current, the higher the voltage barrier across the inductor would become - ***!!!*** Inductors resist change in current by storing energy in a magnetic field. Does not resist the current itself (resistor) but rather *changes in current* ***!!!*** - Current of an inductor must be continuous and cannot change abruptly because a sudden jump in its current would result in an infinite voltage drop, impossible to achieve, so $i_{L}(t^-)=i_{L}(t^+)$ - Voltage across can however change instantaneously to compensate for changes happening in the circuit - Considering simple circuit containing an inductor under DC conditions, $v_{L}=0$, represented as short circuit - See [[The Bare Essentials of Electrical Engineering.pdf#page=186|Example 5.3.4]] ## Finding the Current through an Inductor > [!derivation] > Rearranging $v_{L}(t)$ we see $di_{L}=\frac{1}{L}v_{L}(t)dt$, then integrating: > $\int _{-\infty}^{i_{L}(t)}di_{L}=\frac{1}{L}\int _{-\infty}^tv_{L}(x)dx\to i_{L}(t)-i_{l}(-\infty)=\frac{1}{L}\int _{-\infty}^tv_{L}(x)dx$ > Assuming the inductor current at $-\infty$ is zero and assuming everything up until $t_{0}$ is constant: > $i_{L}(t)=\frac{1}{L}\int _{-\infty}^tv_{L}(x)dx=\underbrace{ \frac{1}{L}\int _{-\infty}^tv_{L}(x)dx }_{ A }+\underbrace{ \frac{1}{L}\int _{t_{0}}^tv_{L}(x)dx }_{ B }$ > Where $A$ represents the current through the inductor at a time $t_{0}$, indicated by $i_{L}(t_{0})$, known as initial condition. Substituting and rearranging: > $\boxed{ i_{L}(t)=\frac{1}{L}\int _{t_{0}}^tv_{L}(x)dx+i_{L}(t_{0}) }^{ }$ ## Energy Storage > [!Derivation] > Substituting $v_{L}(t)$ into power equation we get $p_{L}(t)=L \frac{di_{L}(t)}{dt}i_{L}(t)$ > Integrating over the period in which the conductor is charging, assuming it begins at $t_{0}:$ > $w_{L}(t)=\int _{-\infty}^tp_{L}(x)dx=L\int _{-\infty}^t \frac{di_{L}(x)}{dx}i_{L}(x)dx$ > $=L\int _{-\infty}^t i_{L}(x)dx=\frac{L}{2}i_{L}(x)^2\Biggr|^t_{-\infty}=\frac{L}{2}(i_{L}^2(t)-i_{L}^2(-\infty))$ > Assuming the inductor starts with zero charge, $\boxed{ w_{L}(t)=\frac{1}{2}Li_{L}^2(t) }^{ }$ - [[The Bare Essentials of Electrical Engineering.pdf#page=192|Example 5.3.7]], [[The Bare Essentials of Electrical Engineering.pdf#page=193|Example 5.3.8]] ## Inductor Connections > [!Derivation] > For inductors in series, take KVL around the circuit yields and substitute $v_{L}(t)=L \frac{di_{L}(t)}{dt}:$ > $v_{s}(t)=L_{1} \frac{di_{L}(t)}{dt}+L_{2} \frac{di_{L}(t)}{dt}+\dots+L_{N} \frac{di_{L}(t)}{dt}$ > $=(L_{1}+L_{2}+\dots+L_{N}) \frac{di_{L}(t)}{dt}=L_{eq} \frac{di_{L}(t)}{dt}\to \boxed{ L_{eq}=\sum^N _{n=1}L_{n} }^{ }$ > [!Derivation] > For inductors in parallel, substitute $i_{L}(t)=\frac{1}{L}\int _{t_{0}} ^tv_{L}(x)dx+i_{L}(t_{0})$ into KCL equation: > $i_{L}(t)=\frac{1}{L_{1}}\int _{t_{0}}^tv_{s}(x)dx+i_{L_{1}}(t_{0})+\frac{1}{L_{2}}\int _{t_{0}} ^tv_{s}(x)dx+i_{L_{2}}(t_{0})+\dots+\frac{1}{L_{N} }\int _{t_{0}}^t v_{s}(x)dx+i_{L_{N}}(t_{0})$ > The voltage across a parallel connection is the same: > $i_{L}(t)=\left( \frac{1}{L_{1}}+\frac{1}{L_{2}}+\dots+\frac{1}{L_{N}} \right)\int _{t_{0}}^t v_{s}(x)dx+i_{L_{1}}(t_{0})+i_{L_{2}}(t_{0})+\dots+i_{L_{N}}(t_{0})$ > $=\frac{1}{L_{eq}}\int _{t_{0}}^t v_{s}(x)dx+i_{L_{eq}}(t_{0})\to \boxed{ \frac{1}{L_{eq}}=\sum^N _{n=1} \frac{1}{L_{n}} }^{ }$ # Exercises ## [[The Bare Essentials of Electrical Engineering.pdf#page=200|5.9]] First remove the capacitor branch because it's turned into an open circuit Then use source transformations to get the circuit described (L2R): $60V$ source in series w/ $60\Omega$ resistor, then $V_{C}$ branch, in series w/ $7.5\Omega$ resistor then $5.25V$ source with downward polarity Perform KVL on the entire loop: $60+5.25-60I-7.5I=0\to I=\frac{29}{30}=.9\bar{6}$ Voltage drop occurs between $60V$ source and $V_{C}$ with $60\Omega$ resistor between them: $V_{C}=60-60I=60-60 \frac{29}{30}=\boxed{ 2V }^{ }$