Alternating Current Concept Page - 3

Definition

Explain the charging and discharging of a capacitor on application of AC

1. 0-1: ENERGY IS ABSORBED FROM THE SOURCE DURING THIS QUARTER CYCLE AS THE CAPACITOR IS CHARGED.
2. 1-2: THE ENERGY ABSORBED DURING THE 14${\displaystyle \frac{1}{4}}$ CYCLE 0-1 IS RETURNED DURING THIS QUARTER.
3. 2-3: The capacitor ABSORBS ENERGY during this 14${\displaystyle \frac{1}{4}}$ cycle.
4. 3-4: ENERGY ABSORBED DURING 2-3 IS RETURNED TO THE SOURCE. NET ENERGY ABSORBED IS ZERO.

Formula

Current through a capacitor when AC voltage is applied

i=imsin(ωt+π2)$i=i_{m}sin(\omega t+\frac{\pi }{2})$ 
Where 
i$i$ is the current through the capacitor,
im=ωCvm$i_m=\omega C{v}_m$ is the amplitude of the oscillating current
ω$\omega$ is the angular frequency 
C$C$ is the capacitance
vm$v_m$  is the amplitude of the oscillating voltage

Definition

Current in AC circuit containing resistance and inductance

Let an AC of EMF E be connected to a series combination of a capacitor of pure capacitor C and resistor R.
The source voltage is given by : V=VoSinωt$V=V_{o}Sin\omega t$  
Voltage across resistor is given by : VR=IR$V_R=IR$
Voltage across Inductance is given by :VL=IXL$V_L=I{X}_L$
The phase difference between VR$V_R$ and VL$V_L$ will be 90 Degrees
The resultant voltage will be V2=V2R+V2L$V^2=V{}_R^2+V{}_L^2$
                                                       =IR2+X2L−−−−−−−√$=I\sqrt{R^2+X{}_L^2}$ 
Comparing this with ohms law we get impedance Z
Z=R2+ω2L2−−−−−−−−−√$Z=\sqrt{R^2+{\omega }^2{L}^2}$
Phase difference will be given by ϕ=tan−1ωLR$\varphi =ta{n}^{-1}{\displaystyle \frac{\omega L}{R}}$ 
Current is given by : I=Iosin(ωt−ϕ)$I=I_{o}sin(\omega t-\varphi )$ Where ϕ$\varphi$ is phase difference

Definition

Current across inductor does not change suddenly

Energy stored across inductor is given by E=12LI2$E={\displaystyle \frac{1}{2}}L{I}^2$
Since energy stored in the inductor cannot change suddenly and is directly proportional to I2$I^2$, current across an inductor cannot change suddenly.

Definition

Inductor as an open circuit during switching

The stored energy in an inductor tries to maintain a constant current through its windings. Because of this, inductors oppose changes in current, and act precisely the opposite of capacitors, which oppose changes in voltage. A fully discharged inductor (no magnetic field), having zero current through it, will initially act as an open-circuit when attached to a source of voltage (as it tries to maintain zero current), dropping maximum voltage across its leads.

Example

Inductor as an short circuit

Determine the current, i(t)$i(t)$ , and voltage, v(t)$v(t)$ , for this circuit.
Solution:
This is a dc circuit so the capacitor acts like an open circuit. The capacitor voltage,v(t)$v(t)$, is the voltage across that open circuit. The inductor acts like a short circuit. The inductor current, i(t)$i(t)$ ,is the current in that short circuit. The circuit after replacing the capacitor by an open circuit and replacing the inductor by a short circuit is also given.
Ohms law gives
i(t)=308+4=2.5A$i(t)={\displaystyle \frac{30}{8+4}}=2.5A$
v(t)=4i(t)=4(2.5)=10V$v(t)=4i(t)=4(2.5)=10V$

Formula

Differential equation of a LR circuit for growing current

diϵ−Ri=dtL${\displaystyle \frac{di}{ϵ-Ri}}={\displaystyle \frac{dt}{L}}$

Formula

Differential equation of a LR circuit for decaying current

dii=−RLdt${\displaystyle \frac{di}{i}}=-{\displaystyle \frac{R}{L}}dt$

Definition

Time constant in an LR circuit

The time constant of an LR circuit is the time required to achieve 63% of the maximum current flowing through the circuit. 
Time constant τ=LR$\tau ={\displaystyle \frac{L}{R}}$

Result

General Solution of Differential Equation for an Inductor in LR circuit

Figure shows an inductance L, a resistance R and a source of emf ε$\epsilon$ connected in series through a switch S. Initially, the switch is open and is open and there is no current in the circuit. At t=0$t=0$, the switch is closed and the circuit is completed. As the current increases in the inductor, a self-induced emf (−Ldidt)$(-L\frac{di}{dt})$ is produced. Using Kirchhoffs loop law,
εLdidt=Ri$\epsilon L\frac{di}{dt}=Ri$
Or,  Ldidt=εRi$L\frac{di}{dt}=\epsilon Ri$
Or,  diεRi=dtL$\frac{di}{\epsilon Ri}=\frac{dt}{L}$.
At t=0,i=0$t=0,i=0$ and at time t the current is i. Thus,
∫t0diεRi=∫t0dtL${\displaystyle {\int }_0^t\frac{di}{\epsilon Ri}={\int }_0^t\frac{dt}{L}}$

Or,  −1Rlnε−Riε=tL$-\frac{1}{R}ln\frac{\epsilon -Ri}{\epsilon }=\frac{t}{L}$

Or,  ε−Riε=e−tR/L$\frac{\epsilon -Ri}{\epsilon }=e^{-tR/L}$

Or, ε−Ri=εe−tR/L$\epsilon -Ri=\epsilon e^{-tR/L}$

Or,  i=εR(1−e−tR/L)$i=\frac{\epsilon }{R}(1-e^{-tR/L})$

The constant L / R has dimensions of times and is called the time constant of the LR circuit. Writing L/R = τ$\tau$ and ε/R=i0$\epsilon /R=i_0$, equation becomes
i=i0(1−e−t/τ)$i=i_0(1-e^{-t/\tau })$

Figure shows the plot of the current versus time. The current gradually rises from t = 0 and attains the maximum value i0$i_0$ after long time. At t=τ$t=\tau$, the current is
i=i0(1−1e)=0.63i0

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