Informative line

### Lcr In Series

Practice equations of phase relationship between current and voltage in series LCR circuit, resonance frequency & phasor diagram RLC and impedance of a series LCR circuit.

# Phase Relationship between Current and Voltage in Series LCR Circuit

## Series LCR Circuit

• The circuit in which a resistor, inductor and capacitor are in series is called a series LCR circuit.
• Consider a series LCR circuit connected across an AC voltage source, as shown in figure.
• As inductor, capacitor and resistor are connected in series, the current in all these circuit elements will be same.

### Determination of Phase Relationship

If current is zero at some instant, then the phasor diagrams will be as follows:

### For Purely Resistive Circuit

In resistor, the voltage and current are in same phase.

### For Purely Inductive Circuit

In inductor, the voltage leads the current by 90°.

### For Purely Capacitive Circuit

In capacitor, the voltage lags behind the current by 90

On combining all these phasor diagrams, the following phasor diagram will be obtained :

The resultant phasor diagram of series LCR circuit will be as follows :

Case I : When VL > VC

Case II : When VC > VL

Peak Current, I0 :

Assume that, VL > VC

From right angled triangle ABC,

$$V_0^2=V_R^2+(V_L-V_C)^2$$

$$V_0=\sqrt {V_R^2+(V_L-V_C)^2}$$

$$V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}$$

$$V_0=I_0\sqrt {R^2+(X_L-X_C)^2}$$

$$I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}$$

where,

$$\sqrt {R^2+(X_L-X_C)^2}$$ is known as impedance. It is denoted by Z.

Thus, $$I_0=\dfrac {V_0}{Z}$$

### Determination of phase angle ($$\phi$$)

The phase angle $$\phi$$ between the current and voltage is given as follows:

From right angled triangle ABC,

$$tan\,\phi=\dfrac {V_L-V_C}{V_R}$$

$$\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)$$

### Current in the Circuit

When voltage source $$V = V_0 \,sin \,\omega t$$

$$I = I_0 \sin\, (\omega t –\phi)$$

$$=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)$$

### Condition on Phase Angle

1. When XL>XC, the phase angle is positive, which signifies that the current lags the applied voltage and the circuit becomes inductive.
2. When XL< XC, the phase angle is negative which signifies that the current leads the applied voltage and the circuit becomes capacitive.
3. When XL= XC, the phase angle is zero and the circuit is purely resistive.

#### The voltage in a series LCR circuit is given by V = V0 sin ( $$\omega$$t + $$\pi/3$$ ). If the peak value of current is $$I_0$$, then what can be the function of current?

A For purely resistive circuit, I = I0 sin $$\omega$$t

B For capacitive circuit, I = I0 sin ( $$\omega$$t + $$\pi/6$$ )

C For inductive circuit, I = $$I_0$$ sin ( $$\omega$$t + $$\pi/2$$ )

D For inductive circuit, I = $$I_0$$ sin ( $$\omega$$t -  $$\pi/6$$ )

×

Option (A) is incorrect because in a purely resistive circuit, the voltage and current are in same phase.

But in option (A), voltage is leading current.

Option (B) is incorrect because in a capacitive circuit, the voltage lags behind the current.

But in option (B) current is lagging behind the voltage.

Option (C) is incorrect because in an inductive circuit, the voltage leads the current.

But in option (C) current is leading the voltage.

Option (D) is correct because in an inductive circuit, the voltage leads the current.

### The voltage in a series LCR circuit is given by V = V0 sin ( $$\omega$$t + $$\pi/3$$ ). If the peak value of current is $$I_0$$, then what can be the function of current?

A

For purely resistive circuit, I = I0 sin $$\omega$$t

.

B

For capacitive circuit, I = I0 sin ( $$\omega$$t + $$\pi/6$$ )

C

For inductive circuit, I = $$I_0$$ sin ( $$\omega$$t + $$\pi/2$$ )

D

For inductive circuit, I = $$I_0$$ sin ( $$\omega$$t -  $$\pi/6$$ )

Option D is Correct

# Phasor Diagram for Series LCR Circuit

### Series LCR Circuit

• The circuit in which a resistor, inductor and capacitor are in series is called a series LCR circuit.
• Consider a series LCR circuit connected across an AC voltage source, as shown in figure.
• As inductor, capacitor and resistor are connected in series, the current in all these circuit elements will be same.

## Determination of Phase Relationship

If current is zero at some instant, then the phasor diagrams will be as follows:

### For Purely Resistive Circuit

In resistor, the voltage and current are in same phase.

### For Purely Inductive Circuit

In inductor, the voltage leads the current by 90°.

### For Purely Capacitive Circuit

In capacitor, the voltage lags behind the current by 90°

On combining all these phasor diagrams, the following phasor diagram will be obtained:

The resultant phasor diagram of series LCR circuit will be as follows:

Case I : When VL > VC

Case II : When VC > VL

Peak Current, $$\text I_0$$

Assume that VL > VC

From right angled triangle ABC,

$$V_0^2=V_R^2+(V_L-V_C)^2$$

$$V_0=\sqrt {V_R^2+(V_L-V_C)^2}$$

$$V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}$$

$$V_0=I_0\sqrt {R^2+(X_L-X_C)^2}$$

$$I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}$$

where, $$Z={\sqrt {R^2+(X_L-X_C)^2}}$$ is known as impedance. It is denoted by $$Z$$

Thus, $$I_0=\dfrac{V_0}{Z}$$

### Determination of Phase Angle ($$\phi$$)

The phase angle $$\phi$$ between the current and voltage is given as follows:

From right angled triangle ABC,

$$tan\,\phi=\dfrac {V_L-V_C}{V_R}$$

$$\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)$$

### Current in the Circuit

When voltage source $$V = V_0 \,sin\, \omega t$$

($$I = I_0\, sin\, (\omega t –\phi)$$)

$$I=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)$$

### Condition on Phase Angle

1. When $$X_L > X_C,$$ the phase angle is positive, which signifies that the current lags the applied voltage and the circuit becomes inductive.
2. When $$X_L < X_C$$, the phase angle is negative which signifies that the current leads the applied voltage and the circuit becomes capacitive.
3. When $$X_L = X_C$$, the phase angle is zero and the circuit is purely resistive.

#### Which phasor is incorrect for LCR circuit, if current is zero at some instant?

A

B

C

D

×

Option (A) is incorrect because in purely capacitive circuit, the current phasor leads voltage phasor by 90°.

Option (B) is correct because in purely resistive circuit, the voltage and current are in same phase.

Option (C) is correct because in purely inductive circuit, the voltage leads the current by 90°.

Option (D) is correct because in series  LCR circuit, the current leads capacitive voltage and lags behind inductive voltage by 90°. Also current remains in same phase with resistive voltage.

### Which phasor is incorrect for LCR circuit, if current is zero at some instant?

A
B
C
D

Option A is Correct

# Maximum Voltage across Elements of Series LCR circuit

• Consider a series LCR circuit connected across an AC voltage source, as shown in figure.

### Maximum Voltage across Inductor

$$V_L=I_0\,X_L$$

where,

$$I_0$$  = Peak Current

X= Inductive reactance

### Maximum Voltage across Capacitor

$$V_C=I_0\,X_C$$

where,

$$I_0$$  = Peak Current

XC = Capacitive reactance

### Maximum Voltage across Resistor

$$V_R=I_0\,R$$

where,

$$I_0$$  = Peak Current

R = Resistance of resistor

#### In series LCR circuit, the resistor of resistance $$R=125\,\Omega$$, the capacitor of capacitance $$C=3.50\,\mu F$$ and inductor of inductance L = 1.25 H, are connected across the voltage source of  $$V=100sin120\pi t$$ . Calculate the maximum voltage across each element.

A VR  =30 Volt, VL  = 234.98 Volt, VC  = 20 Volt

B VR  =40 Volt, VL  = 150.72 Volt, VC  = 242.56 Volt

C VR  =20 Volt,  VL  = 30 Volt, VC  = 234.98 Volt

D VR  =60 Volt,  VL  = 90 Volt, VC  = 100 Volt

×

Resistance of resistor, R = 125 $$\Omega$$

Capacitance of capacitor, C = 3.50 $$\mu F$$

Inductance of inductor, L = 1.25 H

Voltage, V = 100 sin 120 $$\pi\,t$$

Inductive reactance, $$X_L=\omega\,L$$

$$X_L=120×3.14×1.25$$

$$X_L=471\,\Omega$$

Capacitive Reactance, $$X_C=\dfrac {1}{\omega\,C}$$

$$X_C=\dfrac {1}{120×3.14×3.50×10^{-6}}$$

$$X_C=758.26\,\Omega$$

$$X_C=758\,\Omega$$

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

$$Z=\sqrt {(125)^2+(471-758)^2}$$

$$Z=\sqrt {(125)^2+(-287)^2}$$

$$Z=\sqrt {15625+82369}$$

$$Z=\sqrt {97994}$$

$$Z=313.03\;\Omega$$

$$Z=313\;\Omega$$

Peak current,  $$I_0 = \dfrac {\text {Peak voltage}}{\text{Impedance}}$$

$$I_0=\dfrac {V_0}{Z}$$

$$I_0=\dfrac {100}{313.03}$$

$$I_0=0.319\;A$$

Maximum Voltage across Resistor

$$V_R=I_0×R$$

$$V_R=0.32×125$$

$$V_R=40\;Volt$$

Maximum Voltage across Inductor

$$V_L=I_0×X_L$$

$$V_L=0.32×471$$

$$V_L=150.72\;Volt$$

Maximum Voltage across Capacitor

$$V_C=I_0×X_C$$

$$V_C=0.32×758$$

$$V_C=242.56\;Volt$$

### In series LCR circuit, the resistor of resistance $$R=125\,\Omega$$, the capacitor of capacitance $$C=3.50\,\mu F$$ and inductor of inductance L = 1.25 H, are connected across the voltage source of  $$V=100sin120\pi t$$ . Calculate the maximum voltage across each element.

A

VR  =30 Volt, VL  = 234.98 Volt, VC  = 20 Volt

.

B

VR  =40 Volt, VL  = 150.72 Volt, VC  = 242.56 Volt

C

VR  =20 Volt,  VL  = 30 Volt, VC  = 234.98 Volt

D

VR  =60 Volt,  VL  = 90 Volt, VC  = 100 Volt

Option B is Correct

# Resonance Frequency of a Series LCR Circuit

### Effect of Frequency on Current in Series LCR Circuit

Inductive reactance, $$X_L=\omega\;L$$

Capacitive reactance, $$X_C=\dfrac {1}{\omega \;C}$$

Current in series LCR circuit is given by  $$I=\dfrac {V}{\sqrt{ R^2+(X_L-X_C)^2}}$$

If frequency is varied, keeping everything else constant, then

• At high frequency, inductive reactance increases, due to which current decreases  $$(X_L\propto\omega)$$.
• At low frequency, capacitive reactance increases, due to which current decreases  $$\left ( X_C\propto\dfrac {1}{\omega} \right)$$

### Resonance Frequency

• If current approaches zero at very low and very high frequencies, there should be some intermediate frequency where, I is maximum.
• Impedance, $$Z={\sqrt{ R^2+(X_L-X_C)^2}}$$
• For maximum current, impedance should be minimum.
• For Z to be minimum,

$$X_L=X_C$$

$$\omega L=\dfrac {1}{\omega C}$$

• The frequency $$\omega _0$$ that satisfies above equation is called the resonance frequency. It is given by $$\omega_0=\dfrac {1} {\sqrt {LC}}$$          {$$\omega=\omega _0$$}

• The relation between current and frequency can be shown with a curve:

$$R_1<R_2<R_3$$

Since, $$I_{max}=\dfrac {V_0}{R}$$

$$I_{max}\propto\dfrac {1}{R}$$

It can be concluded from the graph that at resonance frequency, maximum current is inversely proportional to resistance.

### Graph of Voltage and Current with Time when Frequency of Circuit is less than Resonance Frequency

• When $$\omega<\omega_0$$ , then capacitive reactance increases and circuit becomes more capacitive.
• The current leads the voltage.

### Graph of Voltage and Current with Time when Frequency of Circuit is equal to the Resonance Frequency

• When $$\omega=\omega_0$$, then inductive reactance is equal to capacitive reactance and circuit becomes resistive.
• The current and voltage remain in same phase.

### Graph of Voltage and Current with Time when Frequency of Circuit is more than Resonance Frequency

• When $$\omega>\omega_0$$, the inductive reactance increases and the circuit becomes more inductive.
• The current lags the voltage.

#### Calculate the resonance frequency for series LCR circuit, when resistance of resistor is R = 20 $$\Omega$$, inductance of inductor is L = 5 H and capacitance of capacitor is C = 5 $$\mu F$$.

A 700 s–1

B 100 s–1

C 2000 s–1

D 200 s–1

×

Resistance of resistor, R = 20 $$\Omega$$

Capacitance of capacitor, C = 5 $$\mu\,F$$

Inductance of inductor, L = 5 H

Resonance frequency, $$\omega_0=\dfrac {1}{\sqrt {L\;C}}$$

$$\omega_0=\dfrac {1}{\sqrt {5×5×10^{-6}}}$$

$$\omega_0=\dfrac {1}{5×10^{-3}}=200\;s^{-1}$$

### Calculate the resonance frequency for series LCR circuit, when resistance of resistor is R = 20 $$\Omega$$, inductance of inductor is L = 5 H and capacitance of capacitor is C = 5 $$\mu F$$.

A

700 s–1

.

B

100 s–1

C

2000 s–1

D

200 s–1

Option D is Correct

# Impedance of a Series LCR Circuit

### Series LCR Circuit

• The circuit in which a resistor, inductor and capacitor are in series is called a series LCR circuit.
• Consider a series LCR circuit connected across an AC voltage source, as shown in figure.
• As inductor, capacitor and resistor are connected in series, the current in all these circuit elements will be same.

## Determination of Phase Relationship

If current is zero at some instant, then the phasor diagrams will be as follows:

### For Purely Resistive Circuit

In resistor, the voltage and current are in same phase.

### For Purely Inductive Circuit

In inductor, the voltage leads the current by 90°

### For Purely Capacitive Circuit

In capacitor, the voltage lags behind the current by $$90^\circ$$

On combining all these phasor diagrams, the following phasor diagram will be obtained:

The resultant phasor diagram of series LCR circuit will be as follows:

Case I : When VL > VC

Case II : When VC > VL

Peak Current, $$I_0$$

Assume that VL > VC

From right angled triangle ABC,

$$V_0^2=V_R^2+(V_L-V_C)^2$$

$$V_0=\sqrt {V_R^2+(V_L-V_C)^2}$$

$$V_0=\sqrt {(I_0R)^2+(I_0X_L-I_0X_C)^2}$$

$$V_0=I_0\sqrt {R^2+(X_L-X_C)^2}$$

$$I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}$$

### Determination of phase angle ($$\phi$$)

The phase angle $$\phi$$ between the current and voltage is given as follows:

From right angled triangle ABC,

$$tan\,\phi=\dfrac {V_L-V_C}{V_R}$$

$$\phi=tan^{-1}\left(\dfrac {V_L-V_C}{V_R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {I_0X_L-I_0X_C}{I_0R}\right)$$

$$\phi=tan^{-1}\left(\dfrac {X_L-X_C}{R}\right)$$

### Current in the Circuit

When voltage source $$V = V_0 \sin \omega t$$

$$I = I_0 \,sin\, (\omega t – \phi)$$

$$I=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}\;sin\left ( \omega t-tan^{-1}\dfrac {(X_L-X_C)}{R} \right)$$

### Condition on phase angle

1. When$$X_L > X_C$$, the phase angle is positive, which signifies that the current lags the applied voltage and the circuit becomes more inductive than capacitive.
2. When $$X_L < X_C$$, the phase angle is negative which signifies that the current leads the applied voltage and the circuit becomes more capacitive than inductive.
3. When $$X_L= X_C$$, the phase angle is zero and the circuit is purely resistive.

## Impedance of Series LCR Circuit

• Peak current in series LCR circuit is given by

$$I_0=\dfrac {V_0}{\sqrt {R^2+(X_L-X_C)^2}}$$

where,

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

The quantity Z is known as impedance of series LCR circuit.

### Unit of Impedance

• The unit of impedance is Ohm as unit of R, Xand XC is Ohm.

### Peak Current in terms of Impedance

• Peak current, $$I_0=\dfrac {V_0}{Z}$$

where,

V0 $$\rightarrow$$ Peak Voltage

$$\rightarrow$$ Impedance

#### In series LCR circuit, the resistor of resistance  $$R=125\,\Omega$$ , the capacitor of capacitance $$C=3.50\,\mu F$$  and inductor of inductance L = 1.25 H, are connected across the voltage source of  $$V=100\, sin\, 120\, \pi\, t$$. Calculate the impedance of the circuit.

A $$420\,\Omega$$

B $$300\,\Omega$$

C $$313\,\Omega$$

D $$575.05\,\Omega$$

×

Resistance of resistor, R = 125 $$\Omega$$

Capacitance of capacitor, C = 3.50 $$\mu F$$

Inductance of inductor, L = 1.25 H

Voltage, V = 100 sin 120 $$\pi\,t$$

Angular frequency, $$\omega=120\,\pi$$

Inductive reactance, $$X_L=\omega\,L$$

$$X_L=120×3.14×1.25$$

$$X_L=471\,\Omega$$

Capacitive Reactance, $$X_C=\dfrac {1}{\omega\,C}$$

$$X_C=\dfrac {1}{120×3.14×3.50×10^{-6}}$$

$$X_C=758.26\,\Omega$$

$$X_C=758\,\Omega$$

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

$$Z=\sqrt {(125)^2+(471-758)^2}$$

$$Z=\sqrt {(125)^2+(-287)^2}$$

$$Z=\sqrt {15625+82369}$$

$$Z=\sqrt {97994}$$

$$Z=313.03\;\Omega$$

$$Z=313\;\Omega$$

### In series LCR circuit, the resistor of resistance  $$R=125\,\Omega$$ , the capacitor of capacitance $$C=3.50\,\mu F$$  and inductor of inductance L = 1.25 H, are connected across the voltage source of  $$V=100\, sin\, 120\, \pi\, t$$. Calculate the impedance of the circuit.

A

$$420\,\Omega$$

.

B

$$300\,\Omega$$

C

$$313\,\Omega$$

D

$$575.05\,\Omega$$

Option C is Correct

# RMS Current in a Series LCR Circuit

## Root Mean Square Value of Alternating Current

• The square root of mean square current is called root-mean-square current or rms current.

$$i_{rms}=\sqrt {\overline{i^2}}$$

### Root Mean Square Value

• Root mean square value is a mathematical quantity. It is used to compare alternating current and direct current.
• Root mean square value of the alternating current is the direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.

## Root Mean Square Value of Current in a Series LCR Circuit

$$I_{rms}=\dfrac {V_{rms}}{Z}$$

where,

Z = Impedance of the circuit

$$V_{rms}=$$ rms voltage

#### In series LCR circuit, the resistor of resistance R = 125 $$\Omega$$, the capacitor of capacitance C = 3.50 $$\mu F$$  and inductor of inductance L = 1.25 H are connected across the voltage sources. The rms voltage is $$V_{rms}=313 \,V$$ and angular frequency is $$\omega=120\,\pi$$ rad / sec. Calculate the rms current in the circuit.

A 1 A

B 3 A

C 5 A

D 2 A

×

Resistance of resistor, R = 125 $$\Omega$$

Capacitance of capacitor, C = 3.50 $$\mu F$$

Inductance of inductor, L = 1.25 H

rms Voltage, $$V_{rms} = 313 \,V$$

Inductive reactance, $$X_L=\omega\,L$$

$$X_L=120×3.14×1.25$$

$$X_L=471\,\Omega$$

Capacitive Reactance, $$X_C=\dfrac {1}{\omega\,C}$$

$$X_C=\dfrac {1}{120×3.14×3.50×10^{-6}}$$

$$X_C=758.26\,\Omega$$

$$X_C=758\,\Omega$$

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

$$Z=\sqrt {(125)^2+(471-758)^2}$$

$$Z=\sqrt {(125)^2+(-287)^2}$$

$$Z=\sqrt {15625+82369}$$

$$Z=\sqrt {97994}$$

$$Z=313.03\;\Omega$$

$$Z=313\;\Omega$$

rms current, $$I_{rms}=\dfrac {V_{rms}}{Z}$$

$$I_{rms}=\dfrac {313}{313}=1\;A$$

### In series LCR circuit, the resistor of resistance R = 125 $$\Omega$$, the capacitor of capacitance C = 3.50 $$\mu F$$  and inductor of inductance L = 1.25 H are connected across the voltage sources. The rms voltage is $$V_{rms}=313 \,V$$ and angular frequency is $$\omega=120\,\pi$$ rad / sec. Calculate the rms current in the circuit.

A

1 A

.

B

3 A

C

5 A

D

2 A

Option A is Correct