Alternating Current Concept Page - 6

Example

Problems on LCR circuits excited by an AC source

Example: An LC circuit has L=$=$5 mH and C=$=$20μF$\mu F$. V=5×10−3cosωt$V=5\times {10}^{-3}cos\omega t$ is supplied. ω$\omega$ is twice the resonant frequency. Find the maximum charge stored in the capacitor.

Solution:
L=5mH$L=5mH$ C=20μF$C=20\mu F$
ω=2ωo=2LC−−−√=200010−−√$\omega =2{\omega }_o={\displaystyle \frac{2}{\sqrt{LC}}}=2000\sqrt{10}$
XL=ωL=1010−−√$X_L=\omega L=10\sqrt{10}$
XC=1ωC=2.510−−√$X_C={\displaystyle \frac{1}{\omega C}}=2.5\sqrt{10}$
Z=XL−XC=7.510−−√$Z=X_L-X_C=7.5\sqrt{10}$
Imax=VmaxZ=2.108×10−4A$I_{max}={\displaystyle \frac{V_{max}}{Z}}=2.108\times {10}^{-4}A$
Qmax=CVc=CImaxXc=CI1ωC=Iω$Q_{max}=C{V}_c=C{I}_{max}{X}_c=CI{\displaystyle \frac{1}{\omega C}}={\displaystyle \frac{I}{\omega }}$
=> Qmax=2.108×10−4200010−−√=33.3nC$Q_{max}={\displaystyle \frac{2.108\times {10}^{-4}}{2000\sqrt{10}}}=33.3nC$

Definition

Impedance of an AC circuit

Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. In quantitative terms, it is the complex ratio of the voltage to the current in an alternating current (AC) circuit.
It is written as Z=R2+(XC−XL)2−−−−−−−−−−−−−−√$Z=\sqrt{R^2+{(X_C-X_L)}^2}$
where Z$Z$ is the impedance,
R$R$ is the resistance,
XC$X_C$ is the capacitive reactance,
XL$X_L$ is the inductive reactance.

Definition

Instantaneous current through LCR circuit

i=imsin(ωt+ϕ)$i=i_{m}sin(\omega t+\varphi )$
im=vmZ$i_m=\frac{v_m}{Z}$ ; 
where i$i$ is the instantaneous current,
im$i_m$ is the amplitude of the current,
vm$v_m$ is the amplitude of the voltage,
Z$Z$ is the impedance and
ϕ$\varphi$ is the phase difference between the voltage across the source and the current in the circuit.

Definition

Phasor relations for various circuit components

v2m=v2Rm+(vCm−vLm)2$v_m^2=v_{Rm}^2+{(v_{Cm}-v_{Lm})}^2$
where vm$v_m$ is the amplitude of the resultant potential difference,
vRm=imR$v_{Rm}=i_{m}R$,
vCm=imXC$v_{Cm}=i_m{X}_C$,
vLm=imXL$v_{Lm}=i_m{X}_L$,
im$i_m$ is the magnitude of the alternating current,
R$R$ is the resistance,
XC$X_C$ is the capacitive reactance,
XL$X_L$ is the inductive reactance.

Definition

Impedance diagram

The impedance diagram is a triangle with Z$Z$ as the hypotenuse, and the phase angle ϕ$\varphi$ as a base angle, the base being R$R$ and the other side being XC−XL$X_C-X_L$
If XC>XL$X_C>X_L$, ϕ$\varphi$ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC<XL$X_C<X_L$, ϕ$\varphi$ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.

Example

Phasor-Diagram Solution for an LCR circuit

Example: A series LCR circuit is connected to an a.c. source of variable frequency. Draw a suitable phasor diagram for the amplitude of the current and phase angle.

Solution: The phasor diagram of series RLC circuit is drawn by combining the phasor diagram of resistor, inductor and capacitor. The current flowing in all the elements are same. For drawing the phasor diagram, take current phasor as reference and draw it on horizontal axis as shown in diagram. In case of resistor, both voltage and current are in same phase. So VR$V_R$ and I$I$ are in same direction. In inductor, voltage leads current by 90o${90}^o$ so we draw VL$V_L$ in perpendicular to current phasor in leading direction.Similarly VC$V_C$ is perpendicular to current phasor in lagging direction. In resultant diagram we draw VC$V_C$ in upward direction. Now the resultant, VS$V_S$ which is vector sum of VR$V_R$ and VL−VC$V_L-V_C$.

Formula

Understand the analytical solution (differential equation) for an LCR circuit

The voltage equation for the circuit is:
Ldidt+Ri+qC=v=vmsinωt$L{\displaystyle \frac{di}{dt}}+Ri+{\displaystyle \frac{q}{C}}=v=v_m\sin \omega t$
We know that, i=dqdt.$i={\displaystyle \frac{dq}{dt}}.$ Therefore, didt=d2qdt2${\displaystyle \frac{di}{dt}}={\displaystyle \frac{d^{2}q}{d{t}^2}}$. Thus, in terms of q$q$, the voltage equation becomes:
Ld2qdt2+rdqdt+qC=vmsinωt$L{\displaystyle \frac{d^{2}q}{d{t}^2}}+r{\displaystyle \frac{dq}{dt}}+{\displaystyle \frac{q}{C}}=v_m\sin \omega t$......(i)
Let us assume a solution:
q=qmsin(ωt+ϕ)$q=q_m\sin (\omega t+\varphi )$......(ii)
So that, dqdt=qmωcos(ωt+ϕ)${\displaystyle \frac{dq}{dt}}=q_m\omega \cos (\omega t+\varphi )$......(iii)
and, d2qdt2=−qmω2sin(ωt+ϕ)${\displaystyle \frac{d^{2}q}{d{t}^2}}=-q_m{\omega }^2\sin (\omega t+\varphi )$.....(iv)
On solving (i), (ii), (iii) and (iv), we finally get the current as:
i=dqdt=qmωcos)ωt+θ)=imcos(ωt+θ)$i={\displaystyle \frac{dq}{dt}}=q_m\omega \cos )\omega t+\theta )=i_m\cos (\omega t+\theta )$
or i=imsin(ωt+ϕ)$i=i_m\sin (\omega t+\varphi )$
where im=vmZ=vmR2+(XC−XL)2−−−−−−−−−−−−−−√$i_m={\displaystyle \frac{v_m}{Z}}={\displaystyle \frac{v_m}{\sqrt{R^2+(X_C-X_L{)}^2}}}$
and ϕ=tan−1XC−XLR$\varphi ={\tan }^{-1}{\displaystyle \frac{X_C-X_L}{R}}$

Definition

Current in AC circuit containing capacitance and inductance

Let an AC of EMF E be connected to a series combination of a capacitor of pure capacitor C and inductance L.
The source voltage is given by : V=VoSinωt$V=V_{o}Sin\omega t$  
Voltage across Inductance is given by :VL=IXL$V_L=I{X}_L$
Voltage across Inductance is given by :VC=IXC$V_C=I{X}_C$
The phase difference between VL$V_L$ and VC$V_C$ will be 180 Degrees
As they are oppsite to each other
The resultant voltage will be VL=VC$V_L=V_C$
The impedance of circuit will be zero and current will be infinite in the circuit, It is condition of electrical resonance

BookMarks

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