AC Series and Parallel Circuits

Draw the vector and impedance diagrams of R-L series circuit

1. Circuit Quantities and Phasor Relationships

In a series circuit consisting of a pure resistor (\(R\)) and a pure inductor (\(L\)) connected across an AC supply voltage \(V\), the common loop current \(I\) passes through both elements. Hence, current \(I\) is chosen as the reference phasor along the positive horizontal axis.

• Resistor Voltage Drop (\(V_R\)): In phase with current \(I\) \((V_R = I \cdot R)\).
• Inductor Voltage Drop (\(V_L\)): Leads current \(I\) by \(90^\circ\) \((V_L = I \cdot X_L)\).
• Total Applied Voltage (\(V\)): Phasor sum of \(V_R\) and \(V_L\):

  $$\vec{V} = \vec{V}_R + \vec{V}_L \implies V = \sqrt{V_R^2 + V_L^2}$$

• Phase Angle (\(\phi\)): Supply voltage leads current by an angle \(\phi\) (or current lags voltage by \(\phi\)):

  $$\phi = \tan^{-1}\left(\frac{V_L}{V_R}\right) = \tan^{-1}\left(\frac{X_L}{R}\right)$$

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2. Vector and Impedance Diagrams

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3. Mathematical Formulas from the Diagrams

A. From the Vector Diagram (Voltage Triangle):

• Total Voltage: \(V = \sqrt{V_R^2 + V_L^2}\)
• Power Factor (\(\cos\phi\)):

  $$\cos\phi = \frac{V_R}{V} \quad \text{(lagging)}$$

• Phase Angle: \(\phi = \tan^{-1}\left(\frac{V_L}{V_R}\right)\)

B. From the Impedance Triangle:

Dividing each voltage side of the voltage triangle by the current \(I\) yields the impedance triangle:

• Total Impedance:

  $$Z = \sqrt{R^2 + X_L^2} \quad \Omega$$

• Complex Form:

  $$\vec{Z} = R + jX_L = |Z|\angle\phi$$

• Power Factor:

  $$\cos\phi = \frac{R}{Z} \quad \text{(lagging)}$$

Exam Summary Note: Because inductive reactance causes current to lag voltage, the phasor \(V_L\) and inductive reactance \(X_L\) point vertically upward along \(+90^\circ\) (\(+j\) axis), yielding a lagging power factor.

Definitions in a Parallel AC Circuit: Resonance and Q-Factor

1. Resonance in a Parallel Circuit

Definition: In a parallel AC circuit containing inductive and capacitive branches, resonance (often called anti-resonance) is the steady-state operating condition at which the reactive (quadrature) component of the total supply current becomes zero.

Under this condition, the total input line current is strictly in phase with the applied supply voltage, resulting in an overall unity power factor (\(\cos\phi = 1\)).

Key Conditions and Formulas:

• Susceptance Balance: The total circuit susceptance becomes zero (\(B_{\text{net}} = 0\)), meaning the inductive susceptance balances the capacitive susceptance:

  $$B_L = B_C$$

• Resonant Frequency (\(f_r\)): For a practical parallel resonant circuit consisting of a coil (\(R-L\)) in parallel with a pure capacitor (\(C\)):

  $$f_r = \frac{1}{2\pi} \sqrt{\frac{1}{LC} - \frac{R^2}{L^2}} \quad \text{Hz}$$

  (If the resistance of the coil \(R\) is negligible, \(f_r = \frac{1}{2\pi\sqrt{LC}}\)).
• Dynamic Impedance (\(Z_d\)): At parallel resonance, the circuit impedance reaches its maximum value and is purely resistive:

  $$Z_d = \frac{L}{C \cdot R} \quad \Omega$$

• Minimum Line Current: Because the impedance is at a maximum, the net current drawn from the mains is at its minimum value:

  $$I_{\text{min}} = \frac{V}{Z_d} = \frac{V \cdot C \cdot R}{L}$$

  Due to this characteristic, a parallel resonant circuit is widely referred to as a Rejector Circuit.

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2. Quality Factor (Q-Factor)

Definition: The Quality Factor (Q-Factor) of a parallel resonant circuit is a dimensionless figure of merit that quantifies the sharpness of resonance and frequency selectivity. In parallel resonance, it is defined as the ratio of the circulating current in either reactive branch to the total line current drawn from the supply at resonant frequency.

$$Q = \frac{\text{Circulating Branch Current}}{\text{Total Supply Current at Resonance}} = \frac{I_C}{I_{\text{min}}} \quad \text{or} \quad \frac{I_L}{I_{\text{min}}}$$

Because circulating reactive currents inside the loop are far larger than the line current drawn from the source, the Q-factor in a parallel resonant circuit represents the Current Magnification Factor.

Mathematical Expressions:

• In Terms of Circuit Parameters (\(R, L, C\)):

  $$Q = \frac{\omega_r L}{R} = \frac{2\pi f_r L}{R} \approx \frac{1}{R} \sqrt{\frac{L}{C}}$$

• In Terms of Dynamic Impedance (\(Z_d\)):

  $$Q = \frac{Z_d}{X_L} = \frac{Z_d}{\omega_r L} = \frac{Z_d}{X_C}$$

• In Terms of Bandwidth:

  $$Q = \frac{f_r}{\text{Bandwidth}} = \frac{f_r}{f_2 - f_1}$$

  where \(f_1\) and \(f_2\) are the half-power frequencies.

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Term	Primary Physical Significance	Key Characteristic Formula
Parallel Resonance	Zero net reactive current, unity power factor, minimum line current, maximum impedance.	$$Z_d = \frac{L}{CR}, \quad f_r \approx \frac{1}{2\pi\sqrt{LC}}$$
Q-Factor	Measure of current magnification and sharpness of the tuning response.	$$Q = \frac{I_C}{I} = \frac{1}{R}\sqrt{\frac{L}{C}}$$

Exam Summary Note: While a series resonant circuit causes voltage magnification (\(Q = V_L / V\)), a parallel resonant circuit produces current magnification (\(Q = I_C / I\)).

Vector (Phasor) Diagram of Parallel AC Circuit

Draw the vector diagram of a parallel circuit with one branch consisting of a resistor of 14Ω and a reactance of 20Ω. A second branch consists of a resistor of 25Ω.A potential difference of 100V,50Hz is applied across the combination.  

1. Circuit Specifications & Phasor Values

Taking the common supply voltage across the parallel combination as the reference phasor along the positive horizontal axis:

$$\vec{V} = 100\angle 0^\circ\text{ V}$$

• Branch 1 (Coil: \(R_1 = 14\ \Omega\), \(X_{L1} = 20\ \Omega\)):
  • Impedance: \(Z_1 = \sqrt{14^2 + 20^2} = \sqrt{596} \approx 24.41\ \Omega\)
  • Phase Angle: \(\phi_1 = \tan^{-1}\left(\frac{20}{14}\right) \approx 55.01^\circ \quad (\text{lagging})\)
  • Current: \(I_1 = \frac{100}{24.41} \approx \mathbf{4.10\text{ A}} \implies \vec{I}_1 = 4.10\angle -55.01^\circ\text{ A} = (2.35 - j3.36)\text{ A}\)
• Branch 2 (Non-inductive resistor: \(R_2 = 25\ \Omega\)):
  • Current: \(I_2 = \frac{100}{25} = \mathbf{4.00\text{ A}}\) (strictly in phase with voltage)
  • Phasor: \(\vec{I}_2 = 4.00\angle 0^\circ\text{ A} = (4.00 + j0)\text{ A}\)
• Total Resultant Line Current (\(\vec{I} = \vec{I}_1 + \vec{I}_2\)):
  • \(\vec{I} = (2.35 - j3.36) + (4.00 + j0) = (6.35 - j3.36)\text{ A}\)
  • Magnitude: \(I = \sqrt{(6.35)^2 + (-3.36)^2} \approx \mathbf{7.18\text{ A}}\)
  • Phase Angle: \(\phi = \tan^{-1}\left(\frac{-3.36}{6.35}\right) \approx \mathbf{-27.86^\circ} \quad (\text{lagging})\)

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2. Vector Diagram

R-L Series Circuit Calculations

Problem Statement

A circuit consists of a \(10\ \Omega\) resistance and an \(8\ \Omega\) inductive reactance in series and takes a current of \(6\text{ A}\).

Determine:

1. Voltage across resistance and inductance
2. Total supply voltage
3. Power factor of the circuit

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Given Data

• Resistance (\(R\)) = \(10\ \Omega\)
• Inductive Reactance (\(X_L\)) = \(8\ \Omega\)
• Circuit Current (\(I\)) = \(6\text{ A}\)

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Step-by-Step Solution

(i) Voltage Across Resistance and Inductance

1. Voltage drop across the resistance (\(V_R\)):

$$V_R = I \times R = 6 \times 10 = \mathbf{60\text{ V}}$$

2. Voltage drop across the inductance (\(V_L\)):

$$V_L = I \times X_L = 6 \times 8 = \mathbf{48\text{ V}}$$

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(ii) Total Supply Voltage (\(V\))

In an R-L series circuit, the supply voltage is the phasor sum of \(V_R\) and \(V_L\):

$$V = \sqrt{V_R^2 + V_L^2}$$

$$V = \sqrt{(60)^2 + (48)^2} = \sqrt{3600 + 2304} = \sqrt{5904} \approx \mathbf{76.84\text{ V}}$$

(Alternate method using total impedance: \(Z = \sqrt{R^2 + X_L^2} = \sqrt{10^2 + 8^2} = \sqrt{164} \approx 12.81\ \Omega \implies V = I \times Z = 6 \times 12.81 \approx 76.84\text{ V}\))

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(iii) Power Factor of the Circuit (\(\cos\phi\))

The power factor is given by the ratio of resistance to impedance (or \(V_R\) to \(V\)):

$$\cos\phi = \frac{R}{Z} = \frac{V_R}{V}$$

$$\cos\phi = \frac{60}{76.84} \approx \mathbf{0.781} \quad \text{(lagging)}$$

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Final Results:

1. Voltage across resistance (\(V_R\)): \(60\text{ V}\)
   Voltage across inductance (\(V_L\)): \(48\text{ V}\)
2. Total supply voltage (\(V\)): \(76.84\text{ V}\)
3. Power factor (\(\cos\phi\)): \(0.781\text{ (lagging)}\)

Define and derive Resonance and Resonant Frequency in an R-L-C Series Circuit

1. Definition of Resonant Frequency

In an AC series circuit containing resistance (\(R\)), inductance (\(L\)), and capacitance (\(C\)), electrical resonance occurs when the inductive reactance equals the capacitive reactance, making the total circuit net reactance zero.

The specific supply frequency at which this condition occurs is called the Resonant Frequency (\(f_r\)). At this frequency, the circuit acts as a purely resistive load, the current is in phase with the applied voltage (\(\phi = 0^\circ\)), the power factor is unity (\(\cos\phi = 1\)), the impedance drops to its minimum value (\(Z = R\)), and the current reaches its maximum amplitude.

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2. Graphical Representation

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3. Step-by-Step Derivation of Resonant Frequency

Step 1: Total Impedance Expression

The total impedance \(Z\) of an R-L-C series circuit connected to an AC source of angular frequency \(\omega\) is:

$$Z = R + j(X_L - X_C)$$

where:

• \(X_L = \omega L = 2\pi f L\) = Inductive Reactance
• \(X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}\) = Capacitive Reactance

The magnitude of the impedance is:

$$|Z| = \sqrt{R^2 + (X_L - X_C)^2}$$

Step 2: Condition for Resonance

At resonance, the net reactive component vanishes completely. Therefore, the imaginary part of the impedance must equal zero:

$$X_L - X_C = 0 \implies X_L = X_C$$

Step 3: Equating Reactances

Let \(\omega_r\) be the resonant angular frequency in \(\text{rad/s}\):

$$\omega_r L = \frac{1}{\omega_r C}$$

$$\omega_r^2 = \frac{1}{L C}$$

$$\omega_r = \frac{1}{\sqrt{L C}} \quad \text{rad/s}$$

Step 4: Expressing in Hertz (\(\text{Hz}\))

Since \(\omega_r = 2\pi f_r\):

$$2\pi f_r = \frac{1}{\sqrt{L C}}$$

$$f_r = \frac{1}{2\pi \sqrt{L C}} \quad \text{Hertz (Hz)}$$

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4. Summary of Circuit Characteristics at Resonance

Parameter	Condition at Series Resonance
Net Reactance (\(X\))	Zero (\(X_L = X_C \implies X = 0\))
Impedance (\(Z\))	Minimum and purely resistive (\(Z_{\text{min}} = R\))
Circuit Current (\(I\))	Maximum (\(I_{\text{max}} = \frac{V}{R}\))
Phase Difference (\(\phi\))	Zero (\(0^\circ\)), voltage and current are in phase
Power Factor (\(\cos\phi\))	Unity (\(\cos\phi = 1\))
Circuit Classification	Acceptor Circuit (draws maximum current at \(f_r\))

Exam Key Takeaway: Because the impedance is at a minimum and current is at a maximum at \(f_r\), a series resonant circuit is known as an Acceptor Circuit. The reactive voltages \(V_L\) and \(V_C\) can be many times greater than the supply voltage \(V\), a phenomenon termed voltage magnification.

Write the Procedure of Vector (Phasor) Method in Parallel AC Circuits

The Vector (Phasor) Method is a graphical or component-resolution approach used to find the total current, overall power factor, and power consumed in a circuit having two or more parallel branches connected across a common alternating voltage.

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Phasor Representation and Component Resolution

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Step-by-Step Procedure

Step 1: Choose the Reference Phasor

• Because the supply voltage (\(V\)) is identical and common across all parallel branches, voltage \(V\) is chosen as the reference phasor along the positive X-axis:

  $$\vec{V} = V\angle 0^\circ = V + j0$$

Step 2: Calculate the Impedance and Phase Angle of Each Branch

• For each individual branch \(k\) (where \(k = 1, 2, 3, \dots\)), compute its branch impedance magnitude (\(Z_k\)):

  $$Z_k = \sqrt{R_k^2 + X_k^2}$$

• Determine the phase angle (\(\phi_k\)) between branch current and the supply voltage:

  $$\phi_k = \tan^{-1}\left(\frac{X_k}{R_k}\right)$$

  Note: \(\phi_k\) is lagging (negative angle) for inductive branches, and leading (positive angle) for capacitive branches.

Step 3: Calculate the Magnitude of Branch Currents

• Using Ohm's law, find the current magnitude in each branch:

  $$I_1 = \frac{V}{Z_1}, \quad I_2 = \frac{V}{Z_2}, \quad \dots \quad I_k = \frac{V}{Z_k}$$

Step 4: Resolve Branch Currents into Rectangular Components

Each branch current is resolved into two mutually perpendicular components:

1. Active (In-Phase or Real) Components along the X-axis (\(I_x\)):

   $$\sum I_x = I_1\cos\phi_1 + I_2\cos\phi_2 + \dots$$

2. Reactive (Quadrature or Imaginary) Components along the Y-axis (\(I_y\)):

   $$\sum I_y = \pm I_1\sin\phi_1 \pm I_2\sin\phi_2 \pm \dots$$

   Rule of signs: Take \(+I\sin\phi\) for leading (capacitive) currents and \(-I\sin\phi\) for lagging (inductive) currents.

Step 5: Determine the Total Resultant Current (\(I\))

• The total circuit line current magnitude is the vector sum:

  $$I = \sqrt{\left(\sum I_x\right)^2 + \left(\sum I_y\right)^2}$$

Step 6: Determine the Overall Circuit Phase Angle and Power Factor

• The overall phase angle of the complete parallel combination is:

  $$\phi = \tan^{-1}\left(\frac{\sum I_y}{\sum I_x}\right)$$

• The total power factor (\(\cos\phi\)) of the parallel circuit is:

  $$\cos\phi = \frac{\sum I_x}{I} = \frac{\text{Total Active Component}}{\text{Total Current}}$$

  • If \(\sum I_y\) is negative, the overall power factor is lagging.
  • If \(\sum I_y\) is positive, the overall power factor is leading.

Step 7: Compute Active, Reactive, and Apparent Power

• Active Power (True Power): \(P = V \cdot I \cos\phi = V \cdot \left(\sum I_x\right) \quad [\text{Watts}]\)
• Reactive Power: \(Q = V \cdot I \sin\phi = V \cdot \left|\sum I_y\right| \quad [\text{VAR}]\)
• Apparent Power: \(S = V \cdot I \quad [\text{VA}]\)

Summary Checklist for Exam Problems:

1. Fix \(\vec{V}\) as reference on horizontal axis.
2. Find \(Z_1, Z_2\) and branch currents \(I_1, I_2\).
3. Resolve into horizontal components (\(I\cos\phi\)) and vertical components (\(I\sin\phi\)).
4. Sum components algebraically: \(I = \sqrt{(\sum I_x)^2 + (\sum I_y)^2}\).
5. Compute \(\cos\phi = \frac{\sum I_x}{I}\).

Definitions in a Parallel AC Circuit: Admittance and Resonance

(a) Admittance (\(Y\))

Definition: Admittance is defined as the measure of the ease with which an alternating current is allowed to flow through an AC circuit. It is the reciprocal (inverse) of impedance (\(Z\)).

Mathematically:

$$Y = \frac{1}{Z} \quad \text{Siemens (S) or Mho } (\mho)$$

In complex rectangular form, admittance consists of two components:

$$Y = G \pm jB$$

where:

• \(G\) = Conductance: The real part of admittance, representing the ease with which active current flows through the resistive part (\(G = \frac{R}{Z^2}\)).
• \(B\) = Susceptance: The imaginary part of admittance, representing the ease with which reactive current flows through the reactive part (\(B = \frac{X}{Z^2}\)).
  • Inductive susceptance: \(B_L = \frac{X_L}{Z^2}\) (considered negative, \(-jB_L\))
  • Capacitive susceptance: \(B_C = \frac{X_C}{Z^2}\) (considered positive, \(+jB_C\))

The total magnitude of admittance is:

$$|Y| = \sqrt{G^2 + B^2}$$

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(b) Resonance (Parallel Resonance)

Definition: In a parallel AC circuit, resonance is the operating condition at which the reactive (quadrature) component of the total line current becomes zero, making the circuit entirely resistive. At this condition, the total supply voltage and the resultant line current are completely in phase with each other, resulting in a unity power factor (\(\cos\phi = 1\)).

Key Characteristics of Parallel Resonance (Anti-Resonance):

• Susceptance Balance: The net susceptance of the parallel combination becomes zero (\(B = 0\)), meaning the inductive susceptance balances the capacitive susceptance.
• Minimum Line Current: The total input line current reaches its minimum value:

  $$I_{\text{min}} = \frac{V}{Z_d} = V \cdot G$$

• Maximum Dynamic Impedance (\(Z_d\)): The total effective impedance across the circuit terminals reaches its maximum value:

  $$Z_d = \frac{L}{C \cdot R}$$

• Resonant Frequency (\(f_r\)): For a practical parallel circuit having an inductive coil (\(R\)-\(L\)) in parallel with a pure capacitor (\(C\)), the resonant frequency is:

  $$f_r = \frac{1}{2\pi} \sqrt{\frac{1}{L C} - \frac{R^2}{L^2}} \quad \text{Hz}$$

  (If coil resistance \(R\) is neglected: \(f_r = \frac{1}{2\pi\sqrt{LC}}\)).

Exam Summary Note: A parallel resonant circuit rejects line current at the resonant frequency, which is why it is often called a Rejector Circuit or Anti-Resonant Circuit (unlike a series resonant circuit, which is an Acceptor Circuit).

Phasor (Vector) and Impedance Diagrams of R-L-C Series Circuit (\(X_L > X_C\))

1. Circuit Conditions (\(X_L > X_C\))

In an R-L-C series circuit carrying a sinusoidal current \(I\):

• The current \(I\) is common to all components and is taken as the reference phasor.
• Voltage across resistor: \(V_R = I \cdot R\) (in phase with current \(I\)).
• Voltage across inductor: \(V_L = I \cdot X_L\) (leads current \(I\) by \(90^\circ\)).
• Voltage across capacitor: \(V_C = I \cdot X_C\) (lags current \(I\) by \(90^\circ\)).

When \(X_L > X_C\) (or \(V_L > V_C\)), the circuit behaves as an inductive circuit. The net reactive voltage is \((V_L - V_C)\) acting in the positive vertical direction, and the total voltage \(V\) leads the current \(I\) by a phase angle \(\phi\).

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2. Vector and Impedance Diagrams

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3. Mathematical Equations

1. Total Supply Voltage (\(V\)):

$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$

2. Total Impedance (\(Z\)):

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

3. Phase Angle (\(\phi\)):

$$\phi = \tan^{-1}\left(\frac{V_L - V_C}{V_R}\right) = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$

4. Power Factor (\(\cos\phi\)):

$$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}} \quad \text{(Lagging)}$$

Exam Summary Note: Since \(X_L > X_C\), inductive reactance dominates capacitive reactance. The circuit behaves inductively, resulting in a lagging power factor with the applied voltage leading the current by angle \(\phi\).

Parallel AC Circuit: Current Calculation

Problem Statement

A coil of resistance \(14\ \Omega\) and reactance \(20\ \Omega\) is shunted by a non-inductive resistance of \(25\ \Omega\). A potential difference of \(100\text{ V}\) at \(50\text{ Hz}\) is impressed across the combination.

Find:

1. The current in each branch
2. The total current taken by the combination

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Given Data

• Applied Voltage (\(V\)) = \(100\text{ V}\) (taken as reference: \(\vec{V} = 100\angle 0^\circ\text{ V} = 100 + j0\text{ V}\))
• Supply frequency (\(f\)) = \(50\text{ Hz}\)
• Branch 1 (Coil):
  • Resistance (\(R_1\)) = \(14\ \Omega\)
  • Inductive Reactance (\(X_{L1}\)) = \(20\ \Omega\)
  • Impedance (\(\vec{Z}_1\)) = \(14 + j20\ \Omega\)
• Branch 2 (Non-inductive resistor):
  • Resistance (\(R_2\)) = \(25\ \Omega\)
  • Reactance (\(X_2\)) = \(0\ \Omega\)
  • Impedance (\(\vec{Z}_2\)) = \(25 + j0\ \Omega\)

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Step 1: Impedance and Current in Each Branch

A. Branch 1 (Coil)

Magnitude of impedance of the coil (\(Z_1\)):

$$Z_1 = \sqrt{R_1^2 + X_{L1}^2} = \sqrt{14^2 + 20^2} = \sqrt{196 + 400} = \sqrt{596} \approx \mathbf{24.41\ \Omega}$$

Phase angle of Branch 1 (\(\phi_1\)):

$$\phi_1 = \tan^{-1}\left(\frac{X_{L1}}{R_1}\right) = \tan^{-1}\left(\frac{20}{14}\right) \approx 55.01^\circ \quad (\text{lagging})$$

Current in Branch 1 (\(I_1\)):

$$I_1 = \frac{V}{Z_1} = \frac{100}{24.41} \approx \mathbf{4.10\text{ A}}$$

In polar and rectangular form:

$$\vec{I}_1 = 4.10\angle -55.01^\circ\text{ A}$$

$$\vec{I}_1 = 4.10\cos(55.01^\circ) - j4.10\sin(55.01^\circ) = \mathbf{2.35 - j3.36\text{ A}}$$

B. Branch 2 (Pure Resistor)

Impedance of Branch 2 (\(Z_2\)):

$$Z_2 = R_2 = \mathbf{25\ \Omega}$$

Current in Branch 2 (\(I_2\)):

$$I_2 = \frac{V}{R_2} = \frac{100}{25} = \mathbf{4.00\text{ A}}$$

Since it is purely resistive, \(\vec{I}_2\) is in phase with voltage:

$$\vec{I}_2 = 4\angle 0^\circ\text{ A} = \mathbf{4.00 + j0\text{ A}}$$

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Step 2: Total Circuit Current (\(I\))

The total current is the phasor sum of the branch currents:

$$\vec{I} = \vec{I}_1 + \vec{I}_2$$

$$\vec{I} = (2.35 - j3.36) + (4.00 + j0) = (2.35 + 4.00) - j3.36$$

$$\vec{I} = \mathbf{6.35 - j3.36\text{ A}}$$

Magnitude of total current (\(I\)):

$$I = \sqrt{(6.35)^2 + (-3.36)^2} = \sqrt{40.32 + 11.29} = \sqrt{51.61} \approx \mathbf{7.18\text{ A}}$$

Overall phase angle (\(\phi\)):

$$\phi = \tan^{-1}\left(\frac{-3.36}{6.35}\right) \approx -27.87^\circ \quad (\text{lagging})$$

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Final Answers:

1. Current in coil branch (\(I_1\)): \(4.10\text{ A}\) (lagging the voltage by \(55.01^\circ\))
2. Current in resistor branch (\(I_2\)): \(4.00\text{ A}\) (in phase with voltage)
3. Total circuit current (\(I\)): \(7.18\text{ A}\) (lagging the voltage by \(27.87^\circ\))

Draw the Vector (Phasor) and Impedance Diagrams of an R-C Series Circuit

1. Circuit Quantities

In an R-C series circuit carrying an alternating current \(I\):

• The current \(I\) is common to both elements and is chosen as the reference phasor.
• Voltage drop across the resistor: \(V_R = I \cdot R\) (in phase with \(I\)).
• Voltage drop across the capacitor: \(V_C = I \cdot X_C\) (lags current \(I\) by \(90^\circ\)).
• Total supply voltage: \(V = \sqrt{V_R^2 + V_C^2}\).

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2. Vector (Phasor) & Impedance Diagrams

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3. Equation of Power Factor

The power factor (\(\cos\phi\)) of an R-C series circuit is the cosine of the phase angle between the total voltage and the total current.

From the impedance triangle and voltage triangle:

$$\text{Power Factor } (\cos\phi) = \frac{V_R}{V} = \frac{R}{Z}$$

Derivation of Active Power in an R-L Series Circuit

1. Circuit Quantities and Phasor Relationship

Consider an alternating sinusoidal voltage applied across a series combination of resistance (\(R\)) and inductance (\(L\)).

Let the instantaneous supply voltage be taken as the reference phasor:

$$v(t) = V_m \sin(\omega t)$$

In an inductive circuit, the circuit current lags behind the applied voltage by a phase angle \(\phi\) (where \(0 < \phi < 90^\circ\)). Therefore, the instantaneous current is:

$$i(t) = I_m \sin(\omega t - \phi)$$

where:

• \(V_m\) = Maximum (peak) value of voltage
• \(I_m\) = Maximum (peak) value of current
• \(\omega\) = Angular frequency in \(\text{rad/s}\)
• \(\phi = \tan^{-1}\left(\frac{X_L}{R}\right)\) = Phase angle of the circuit

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2. Step-by-Step Derivation

Step 1: Instantaneous Power (\(p\))

The instantaneous power supplied to the circuit at any instant is the product of instantaneous voltage and instantaneous current:

$$p = v(t) \cdot i(t) = [V_m \sin(\omega t)] \cdot [I_m \sin(\omega t - \phi)]$$

$$p = V_m I_m \sin(\omega t) \sin(\omega t - \phi)$$

Step 2: Applying Trigonometric Identity

Using the identity \(2 \sin A \sin B = \cos(A - B) - \cos(A + B)\):

$$p = \frac{V_m I_m}{2} \Big[ \cos\big(\omega t - (\omega t - \phi)\big) - \cos\big(\omega t + (\omega t - \phi)\big) \Big]$$

$$p = \frac{V_m I_m}{2} \big[ \cos(\phi) - \cos(2\omega t - \phi) \big]$$

$$p = \frac{V_m I_m}{2} \cos(\phi) - \frac{V_m I_m}{2} \cos(2\omega t - \phi)$$

Step 3: Average (Active) Power over a Complete Cycle (\(P\))

The total average power (Active Power, \(P\)) taken by the circuit over one complete cycle (\(T = \frac{2\pi}{\omega}\)) is the time-average of the instantaneous power:

$$P = \frac{1}{2\pi} \int_{0}^{2\pi} p \, d(\omega t)$$

$$P = \frac{1}{2\pi} \int_{0}^{2\pi} \left[ \frac{V_m I_m}{2} \cos(\phi) - \frac{V_m I_m}{2} \cos(2\omega t - \phi) \right] d(\omega t)$$

Since the second term \(\cos(2\omega t - \phi)\) is a double-frequency sinusoidal component, its average value over a full period is zero:

$$\frac{1}{2\pi} \int_{0}^{2\pi} \cos(2\omega t - \phi) \, d(\omega t) = 0$$

Thus, only the constant first term remains:

$$P = \frac{V_m I_m}{2} \cos(\phi)$$

Step 4: Expressing in RMS Values

Writing \(\frac{V_m I_m}{2}\) in terms of RMS quantities (\(V = \frac{V_m}{\sqrt{2}}\) and \(I = \frac{I_m}{\sqrt{2}}\)):

$$P = \left(\frac{V_m}{\sqrt{2}}\right) \left(\frac{I_m}{\sqrt{2}}\right) \cos(\phi)$$

$$P = V I \cos(\phi) \quad \text{Watts (W)}$$

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3. Alternate Expressions of Active Power

• In Terms of Resistance (\(R\)): From the impedance triangle, the power factor is \(\cos(\phi) = \frac{R}{Z}\), and the applied voltage is \(V = I \cdot Z\). Substituting these:

  $$P = (I \cdot Z) \cdot I \cdot \left(\frac{R}{Z}\right) = I^2 R \quad \text{Watts}$$

  (This proves that in an R-L series circuit, active power is consumed exclusively by the resistance; the pure inductor consumes zero average active power).

Key Summary:

• Active Power (True Power): \(P = V I \cos\phi = I^2 R\) [Unit: Watts / kW]
• \(\cos\phi\): Power Factor of the circuit (\(\cos\phi = \frac{R}{Z}\), lagging)
• Inductor Power: Pure inductance consumes no net active power over a complete cycle.