# Practical Electronics/Parallel RC
## Parallel RC
### Circuit Impedance

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

### Circuit Response

    $I=I_{R}+I_{C}$
    $I={\frac {V}{R}}+C{\frac {dV}{dt}}$
    $V=(IR-RC{\frac {dV}{dt}})$

## Parallel RL
### Circuit Impedance

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

### Circuit Response

    $I=I_{R}+I_{L}$
    $I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt$
    $V=IR-{\frac {R}{L}}\int Vdt$

## Parallel LC
### Circuit Impedance

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

### Circuit response

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

## Parallel RLC
### Circuit Impedance

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

### Circuit response

    $I=I_{R}+I_{L}+I_{C}$
    $I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}$
    $I={\frac {V}{R}}+{\frac {1}{L}}\int Vdt+C{\frac {dV}{dt}}$
    $V=IR-{\frac {R}{L}}\int Vdt-CR{\frac {dV}{dt}}$

#### Natural Respond

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

#### Forced Respond

    $I_{t}=IR+L{\frac {dI}{dt}}+{\frac {1}{C}}\int Idt$
Second ordered equation that has two roots

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

Where

   ##### $\alpha ={\frac {R}{2L}}$
   ##### $\beta ={\frac {1}{\sqrt {LC}}}$

The current of the network is given by

   ##### A e$^{ω1t}$ + B e$^{ω2t}$

From above
    When ${\alpha ^{2}=\beta ^{2}}$, there is only one real root
    ω = -α
    When ${\alpha ^{2}>\beta ^{2}}$, there are two real roots
    ω = -α ± ${\sqrt {\alpha ^{2}-\beta ^{2}}}$
    When ${\alpha ^{2}<\beta ^{2}}$, there are two complex roots
    ω = -α ± j${\sqrt {\beta ^{2}-\alpha ^{2}}}$

#### Resonance Response

At resonance, the impedance of the frequency dependent components cancel out . Therefore the net voltage of the circui is zero
$Z_{L}-Z_{C}=0$ and $V_{L}+V_{C}=0$
    $\omega L={\frac {1}{\omega C}}$
    $\omega ={\sqrt {\frac {1}{LC}}}$
    $Z=Z_{R}+(Z_{L}-Z_{C})=Z_{R}=R$
    $I={\frac {V}{R}}$
At Resonance Frequency
    $\omega ={\sqrt {\frac {1}{LC}}}$ .
    $I={\frac {V}{R}}$ . 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 , ${\sqrt {\frac {1}{LC}}}$ , 00 we have the plot of I versus ω . From the plot If current is reduced to halved of the value of peak current $I={\frac {V}{2R}}$ , this current value is stable over a Frequency Band ω$_1$ - ω$_2$ where ω$_1$ = ω$_o$ - Δω, ω$_2$ = 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 $\omega ={\sqrt {\frac {1}{LC}}}$ . Current is at its maximum value $I={\frac {V}{R}}$ . Then, adjust the value of R to have a value less than the peak current $I={\frac {V}{R}}$ by increasing R to have a desired frequency band .
* If R is increased from R to 2R then the current now is $I={\frac {V}{2R}}$ which is stable over a band of frequency
    ω$_1$ - ω$_2$ where
       ω$_1$ = ω$_o$ - Δω
       ω$_2$ = ω$_o$ + Δω
For value of I < $I={\frac {V}{2R}}$ . The circuit respond to Wide Band of frequencies . For value of $I={\frac {V}{R}}$ < I > $I={\frac {V}{2R}}$ . The circuit respond to Narrow Band of frequencies

## Summary

| Circuit | Symbol | Series | Parallel |
| RC | | | A parallel RC Circuit |
| Impedance | Z | $Z_{t}=R+{\frac {1}{\omega C}}={\frac {\omega CR+1}{\omega C}}$ | ${\frac {1}{Z_{t}}}={\frac {1}{Z_{R}}}+{\frac {1}{Z_{C}}}={\frac {1}{R}}+\omega C={\frac {R}{\omega CR+1}}$ |
| Frequency | $\omega _{o}=2f_{o}$ | $Z_{R}=Z_{C}$ $R={\frac {1}{\omega C}}$ $\omega ={\frac {1}{CR}}$ | ${\frac {1}{R}}={\frac {1}{\omega C}}$ ${\frac {1}{R}}=\omega C$ $\omega ={\frac {1}{CR}}$ |
| Voltage | V | $V=IR+{\frac {1}{C}}\int Idt$ | $I={\frac {V}{R}}+C{\frac {dV}{dt}}$ |
| Current | I | $\int Idt=C(V-IR)$ | ${\frac {dV}{dt}}={\frac {1}{C}}(I-{\frac {V}{R}})$ |
| 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 |