RC Circuit
An RC circuit is a basic electrical setup where a resistor (R) and a capacitor (C) are connected together. These circuits are important for controlling how voltage changes over time. Whether you're charging or discharging a capacitor, the RC circuit follows a clear pattern based on its time constant (RC). By using simple RC circuit formulas, you can calculate how quickly voltage builds up across the capacitor or drops when the power is turned off. These circuits are used in timing, filtering signals, and smoothing voltage in real-world electronics.
RC Series Circuit Diagram
Voltage and current relation
Voltage across resistor: $V_R = IR$ (in phase with current)
Voltage across inductor: $V_C = I X_C$ (lags current by $90^\circ$)
Phasor Diagram
Taking the current as the reference phasor, the phasor diagram for the circuit can be constructed accordingly. The voltage across the resistor, $V_R=IR$, is in phase with the current and is represented by phasor OA in both magnitude and direction. The voltage across the capacitor, $V_C=IX_C$, lags the current by 90° and is shown by phasor AB. The total or applied voltage VV is obtained by the vector (phasor) sum of these two voltage drops.
$\text{Let current } I \text{ be the reference phasor.}$
$V_R = IR \quad \text{(in phase with } I \text{)}$
$\displaystyle V_C = IX_C = I \cdot \frac{1}{\omega C} \quad \text{(lags } I \text{ by } 90^\circ)$
$\text{Using phasor representation:}$
$\vec{V}_R = IR \angle 0^\circ, \quad \vec{V}_C = IX_C \angle -90^\circ$
$\text{The applied voltage is the phasor sum:}$
$\vec{V} = \vec{V}_R + \vec{V}_C$
$\Rightarrow V = \sqrt{(IR)^2 + (IX_C)^2} = I \sqrt{R^2 + X_C^2}$
$\displaystyle \text{Where } X_C = \frac{1}{\omega C}$
$\displaystyle \therefore V = I \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2}$
Impedance of the RC circuit is:
$\boxed{Z = \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2 }}$
Derivation of Power in an RC Series Circuit
$V(t) = V_m \sin(\omega t)$$i(t) = I_m \sin(\omega t + \phi)$
$\phi$ — Phase angle between voltage and current
$p(t) = v(t) \cdot i(t) = V_m \sin(\omega t) \cdot I_m \sin(\omega t + \phi)$
Using the identity:
$\displaystyle \sin A \cdot \sin B = \frac{1}{2} \left[ \cos(A - B) - \cos(A + B) \right]$
Substitute into the power equation:
$\displaystyle p(t) = V_m I_m \cdot \frac{1}{2} \left[ \cos(\phi) - \cos(2\omega t + \phi) \right]$
$\displaystyle p(t) = \frac{V_m I_m}{2} \cos(\phi) - \frac{V_m I_m}{2} \cos(2\omega t + \phi)$
The average power is the time average of \( p(t) \) over one full cycle:
$\displaystyle P_{\text{avg}} = \frac{V_m I_m}{2} \cos(\phi)-0$
Now, convert peak values to RMS values:
$\displaystyle V_{\text{rms}} = \frac{V_m}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$
$\displaystyle P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi)$
Final Power Formula
$\boxed{P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi)}$
Power Factor
The power factor (PF) is defined as the cosine of the phase angle between the voltage and current:$\text{Power Factor} = \cos\phi$
Interpretation:The capacitor stores and returns energy, but does not consume it.
Therefore, real power is only dissipated by the resistor R.
Phase Angle
$\displaystyle \tan(\phi) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{V_C}{V_R} = \frac{I X_C}{I R} = \frac{X_C}{R}$
So, the phase angle is:
Interpretation
Since voltage across the capacitor lags the current, the total voltage also **lags the current**.Therefore, the **current leads the voltage** by the angle \( \phi \).