In this section, we will study a series combination of a Resistor, an Inductor, and a Capacitor also known as the Series LCR circuit. We will study the growth of the current and other quantities in this circuit. These circuits are the fundamental components of many important devices. Let’s study the fundamental elements of this circuit.

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**Series LCR circuit**

When a constant voltage source or battery is connected across a resistor, current is developed in it. This current has a unique direction and flows from the negative terminal of the battery to its positive terminal. The magnitude of current remains constant as well.

If the direction of current through this resistor changes periodically or alternately, then the current is called **alternating current**. An alternating current or AC generator or AC dynamo can be used as AC voltage source.

The figure shows basic LCR series circuit where a voltage V_{s} is applied across RLC series circuit. We will solve this circuit by using vector method. Â Vector drawn for resistance is along the X-axis because current and voltage are in phase in case of a purely resistive circuit and magnitude will be R.

**You can download Alternating Current Cheat Sheet by clicking on the download button below**

**Browse more Topics under Alternating Current**

- Representation of AC Current and Voltage by Rotating Vectors â€“ Phasors
- AC Voltage Applied to a Resistor
- The AC Voltage Applied to a Capacitor
- AC Voltage Applied to an Inductor
- Power in AC Circuit: The Power Factor
- LC Oscillations
- Transformers

Inductors and capacitors will be represented by their respective reactance. In capacitor current leads the potential by 90Â° hence reactance will be along the positive Y-axis with magnitude 1/Ï‰C and similarly inductive reactance will be along the negative Y-axis with magnitude Ï‰L as current in inductor lacks by 90Â°.

### Net Reactance ‘X’

The resultant of X_{L} and X_{C} in the positive Y – axis can be given as:

\(X = \quad { X }_{ C }-{ X }_{ L }=\left( \frac { 1 }{ \omega C } -\omega L \right) \)

This is the net reactance of the circuit.

### Total Impedance ‘Z’

Now we need to find total impedance of RLC circuit. And it can be given as:

\(Z = \sqrt { { R }^{ 2 }+{ \left( \frac { 1 }{ \omega C } -\omega L \right) }^{ 2 } }Â \)

And this resultant impedance makes angle Î¸ with X-axis which can be give as â€“

\(\tan { \theta } =\frac { \left( \frac { 1 }{ \omega C } -\omega L \right) }{ R }Â \)

### Current in The Circuit

Current in circuit can be given as:

\({ V }_{ S }\quad ={ V }_{ m }\sin { \omega t } \\ { I }_{ S }\quad =\frac { { V }_{ S } }{ Z } \\ { I }_{ S }\quad =\frac { { V }_{ m }\sin { \omega t } }{ \sqrt { { R }^{ 2 }+{ \left( \frac { 1 }{ \omega C } -\omega L \right) }^{ 2 } } } \\ { I }_{ S }\quad ={ I }_{ m }\sin { \omega t }Â \)

Where I_{m} is peak current.

\({ I }_{ m }\quad =\frac { { V }_{ m } }{ \sqrt { { R }^{ 2 }+{ \left( \frac { 1 }{ \omega C } -\omega L \right) }^{ 2 } } } =\frac { { V }_{ m } }{ Z }Â \)

- Case 1. Â X
_{C }> X_{L}

When X_{C }> X_{L}, resultant vector for net reactance will be along positive Y-axis and value of Î¸ will be positive. Hence current will lead voltage and circuit will behave as resistive-capacitive circuit.

- Case 2. Â X
_{C }< X_{L}

When X_{C }> X_{L}, the resultant vector for net reactance will be along negative Y-axis and value of Î¸ will be negative. Hence voltage will lead current and the circuit will behave as a resistive-inductive circuit.

- Case 2. Â X
_{C }= X_{L}

When X_{C }= X_{L}, the resultant vector for net reactance will be zero and value of Î¸ will be zero. Hence current and voltage will be in the same phase and circuit will behave as purely resistive circuit and peak current will be maximum.

### Resonance

Peak current is maximum when:

\(\frac { 1 }{ \omega C } -\omega L=0\\ \frac { 1 }{ \omega C } =\omega L\\ \omega =\frac { 1 }{ \sqrt { LC } } =2\pi f\\ f=\frac { 1 }{ 2\pi \sqrt { LC } }Â \)

This frequency is known as resonance frequency.

## Solved Example For You

Q1: An LCR series circuit with inductance 1mH, capacitance 100mF and resistance 1K Ohm are connected to AC voltage source. Find the frequency of source for which current through the resistor is maximum.

Ans: Maximum current will be drawn in circuit in case of resonance frequency which is given by:

\(f=\frac { 1 }{ 2\pi \sqrt { LC } }Â \)

\(f=\frac { 1 }{ 2\pi \sqrt { 1\times { 10 }^{ -3 }\times 100\times { 10 }^{ -3 } } } =\frac { 100 }{ 2\pi } \simeq 16Hz\)

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