Practical Electronics/Parallel RC

${\displaystyle {\frac {1}{Z}}={\frac {1}{Z_{R}}}+{\frac {1}{Z_{C}}}}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{R}}+j\omega C={\frac {j\omega CR+1}{R}}}$

${\displaystyle Z=R{\frac {1}{j\omega CR+1}}}$

${\displaystyle I=I_{R}+I_{C}}$

${\displaystyle I={\frac {V}{R}}+C{\frac {dV}{dt}}}$

${\displaystyle V=(IR-RC{\frac {dV}{dt}})}$

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${\displaystyle {\frac {1}{Z}}={\frac {1}{Z_{R}}}+{\frac {1}{Z_{L}}}}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{R}}+{\frac {1}{j\omega L}}={\frac {R+j\omega L}{j\omega RL}}}$

${\displaystyle Z={\frac {j\omega RL}{R+j\omega L}}=j\omega L{\frac {1}{1+j\omega {\frac {L}{R}}}}}$

${\displaystyle I=I_{R}+I_{L}}$

${\displaystyle I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt}$

${\displaystyle V=IR-{\frac {R}{L}}\int Vdt}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{Z_{L}}}+{\frac {1}{Z_{C}}}}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{j\omega L}}+j\omega C={\frac {(j\omega )^{2}LC+1}{j\omega L}}}$

${\displaystyle Z={\frac {j\omega L}{(j\omega )^{2}LC+1}}}$

${\displaystyle I=I_{L}+I_{C}}$

${\displaystyle I={\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{Z_{R}}}+{\frac {1}{Z_{L}}}+{\frac {1}{Z_{C}}}}$

${\displaystyle {\frac {1}{Z}}={\frac {1}{R}}+{\frac {1}{j\omega L}}+j\omega C}$

${\displaystyle {\frac {1}{Z}}={\frac {(j\omega )^{2}RLC+j\omega L+R}{j\omega RL}}}$

${\displaystyle {\frac {1}{Z}}={\frac {(j\omega )^{2}LC+j\omega {\frac {L}{R}}+1}{j\omega L}}}$

${\displaystyle I=I_{R}+I_{L}+I_{C}}$

${\displaystyle I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}}$

${\displaystyle I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}}$

${\displaystyle V=IR-{\frac {R}{L}}\int Vdt-CR{\frac {dV}{dt}}}$

${\displaystyle 0={\frac {V}{R}}+{\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}}$

${\displaystyle I_{t}=IR+L{\frac {dI}{dt}}+{\frac {1}{C}}\int Idt}$

Second ordered equation that has two roots

ω = -α ± ${\displaystyle {\sqrt {\alpha ^{2}-\beta ^{2}}}}$

Where

${\displaystyle \alpha ={\frac {R}{2L}}}$

${\displaystyle \beta ={\frac {1}{\sqrt {LC}}}}$

The current of the network is given by

A e^{ω1 t} + B e^{ω2 t}

From above

When ${\displaystyle {\alpha ^{2}=\beta ^{2}}}$, there is only one real root

ω = -α

When ${\displaystyle {\alpha ^{2}>\beta ^{2}}}$, there are two real roots

ω = -α ± ${\displaystyle {\sqrt {\alpha ^{2}-\beta ^{2}}}}$

When ${\displaystyle {\alpha ^{2}<\beta ^{2}}}$, there are two complex roots

ω = -α ± j${\displaystyle {\sqrt {\beta ^{2}-\alpha ^{2}}}}$

At resonance, the impedance of the frequency dependent components cancel out . Therefore the net voltage of the circui is zero

${\displaystyle Z_{L}-Z_{C}=0}$ and ${\displaystyle V_{L}+V_{C}=0}$

${\displaystyle \omega L={\frac {1}{\omega C}}}$

${\displaystyle \omega ={\sqrt {\frac {1}{LC}}}}$

${\displaystyle Z=Z_{R}+(Z_{L}-Z_{C})=Z_{R}=R}$

${\displaystyle I={\frac {V}{R}}}$

At Resonance Frequency

${\displaystyle \omega ={\sqrt {\frac {1}{LC}}}}$ .

${\displaystyle I={\frac {V}{R}}}$ . Current is at its maximum value

Further analyse the circuit

At ω = 0, Capacitor Opened circuit . Therefore, I = 0 .

At ω = 00, Inductor Opened circuit . Therefore, I = 0 .

With the values of Current at three ω = 0 , ${\displaystyle {\sqrt {\frac {1}{LC}}}}$ , 00 we have the plot of I versus ω . From the plot If current is reduced to halved of the value of peak current ${\displaystyle I={\frac {V}{2R}}}$ , this current value is stable over a Frequency Band ω₁ - ω₂ where ω₁ = ω_o - Δω, ω₂ = ω_o + Δω

• In RLC series, it is possible to have a band of frequencies where current is stable, ie. current does not change with frequency . For a wide band of frequencies respond, current must be reduced from it's peak value . The more current is reduced, the wider the bandwidth . Therefore, this network can be used as Tuned Selected Band Pass Filter . If tune either L or C to the resonance frequency ${\displaystyle \omega ={\sqrt {\frac {1}{LC}}}}$ . Current is at its maximum value ${\displaystyle I={\frac {V}{R}}}$ . Then, adjust the value of R to have a value less than the peak current ${\displaystyle I={\frac {V}{R}}}$ by increasing R to have a desired frequency band .

• If R is increased from R to 2R then the current now is ${\displaystyle I={\frac {V}{2R}}}$ which is stable over a band of frequency

ω₁ - ω₂ where

ω₁ = ω_o - Δω

ω₂ = ω_o + Δω

For value of I < ${\displaystyle I={\frac {V}{2R}}}$ . The circuit respond to Wide Band of frequencies . For value of ${\displaystyle I={\frac {V}{R}}}$ < I > ${\displaystyle I={\frac {V}{2R}}}$ . The circuit respond to Narrow Band of frequencies

Circuit	Symbol	Series	Parallel
RC
Impedance	Z	${\displaystyle Z_{t}=R+{\frac {1}{\omega C}}={\frac {\omega CR+1}{\omega C}}}$	${\displaystyle {\frac {1}{Z_{t}}}={\frac {1}{Z_{R}}}+{\frac {1}{Z_{C}}}={\frac {1}{R}}+\omega C={\frac {R}{\omega CR+1}}}$
Frequency	${\displaystyle \omega _{o}=2f_{o}}$	${\displaystyle Z_{R}=Z_{C}}$ ${\displaystyle R={\frac {1}{\omega C}}}$ ${\displaystyle \omega ={\frac {1}{CR}}}$	${\displaystyle {\frac {1}{R}}={\frac {1}{\omega C}}}$ ${\displaystyle {\frac {1}{R}}=\omega C}$ ${\displaystyle \omega ={\frac {1}{CR}}}$
Voltage	V	${\displaystyle V=IR+{\frac {1}{C}}\int Idt}$	${\displaystyle I={\frac {V}{R}}+C{\frac {dV}{dt}}}$
Current	I	${\displaystyle \int Idt=C(V-IR)}$	${\displaystyle {\frac {dV}{dt}}={\frac {1}{C}}(I-{\frac {V}{R}})}$
Phase Angle		Tan θ = 1/2πf RC f = 1/2π Tan CR t = 2π Tan CR	Tan θ = 1/2πf RC f = 1/2π Tan CR t = 2π Tan CR