An RLC series circuit contains all the three passive electrical components, Resistor Capacitor, and Inductor in series across an AC source. As there is only one path for current in a series combination, the current in all these components is the same in magnitude and phase.

We know that voltage and current are in phase in pure resistor while current leads in pure capacitive circuit and in case of pure inductive circuit, current lags. So, what about the voltage across each component, if their behavior is so different?

To answer the question, we will first find the impedance of the series RLC circuit.

Before going further, I would like to take the current phasor as the reference. Because the current is the same in all the components of the circuit and it is more robust to compare different quantities to the common current.

## Impedance Triangle in RLC series circuit:

The total impedance of an RLC series circuit can be found using impedance triangle. First of all, we have to consider the resistance and reactance of each component then put them into the impedance triangle to find the total impedance of the circuit.

Resistance : $R∠0^o$

Inductor Reactance: $X_L=ωL∠90^{o}$

Capacitor Reactance: $X_{ C }=\frac{1}{ωC∠-90^{ o }}$

By observing the above equations, the capacitive reactance and inductive reactance is completely opposite in direction of each other.

In impedance triangle, resistance, capacitive reactance, and inductive reactance phasors will be added to each other using parallelogram law or head and tail rule. The final impedance can be leading or lagging depending upon the difference between the capacitive and inductive reactance magnitude. If the capacitive reactance is larger than inductive reactance, impedance phasor will make a negative angle with the horizontal line or vice versa.

In the above impedance triangle, I have assumed inductive reactance is larger and it is making a positive angle with resistance phasor. The total impedance of the circuit is represented with the following formula

$Z=\sqrt{ (R^{ 2 }+(X_{ L }-X_{ c })^{ 2 }) }$

## Current in RLC series circuit:

According to the Ohms Law in AC circuit, the current of an AC circuit can be found using the following formula

$I=\frac {V}{ Z}$

The phase difference between the current and voltage will depend upon the impedance. If the impedance is more inductive, the current will lag and if the impedance is more capacitive, then current will be leading.

As I have assumed inductive reactance larger than capacitive, so the current, in this case, will lag behind the source voltage.

## Voltage across each component in RLC series circuit :

Applying KVL at the circuit, we will get the following equation

$V_s=V_R+V_L+V_C$

Using the basic current and voltage relationship in resistor, inductor, and capacitor current flow, the above equation can be modified as follow

$V_s=IR+\frac{L di}{dt}+\frac{1}{C}\int {idt}$

The voltage across each circuit element can be found using the following formula

Resistor voltage drop: $V_{ R }=IR$

Inductor voltage drop: $V_{ L}=IωL∠-90^{ o }$

Capacitor voltage drop: $ V_{ C }=\frac{I}{ωC}∠90^{ o }$

The voltage drop in resistor will be in phase with current, in case of the capacitor the current will leads voltage drop and for the inductor, the current of the inductor will be lag from voltage drop in the inductor. The same thing is represented with the phasor diagram.

The above vectors from the above diagram can be added vectorially which will get us the voltage triangle. The vertical component of the triangle shows the voltage drop across reactance (inductive and capacitive) and horizontal component shows drop across the resistance.

## Example wiht solution:

For the given circuit diagram calculate the circuit impedance, current, voltage across each component and power factor. Also draw the phasor diagram of current and voltage, impedance triangle and voltage triangle.

First of all, let me calculate the total impedance with the following formula

Resistance: $R=12\Omega$

Inductive Reactance: $X_{ L }=ωL=2\pi fL=2×\pi×50×0.15=47.1\Omega$

Capacitive Reactance: $X_C=\frac{1}{ωC}=\frac{1}{2\pi fC}=\frac{1}{2×\pi×50×100×10^{-6} }=31.83\Omega$

Now the total impedance will be

$Z=\sqrt{R^{ 2 }+(X_{ L }-X_{ C })^{ 2 }}\\ Z=\sqrt{12^{ 2 }+(47.13-31.83)^{ 2 }}\\ Z=\sqrt{144+234}\\ Z=19.4\Omega$

Where the current is

$I=\frac{V}{Z}=\frac{100}{19.4}=5.14amps$

Voltage across each component in RLC circuit

Voltage across resistor: $V_{ R }=IR=5.14×12=61.7v$

Voltage across capacitor: $V_{ C }=IX_{ C }=5.14×31.8=163.5v$

Voltage across inductor: $V_{ L }=IX_{ L }=5.14×47.13=242.2v$

Observing the above individual voltages, their scaler summation can get us a larger voltage than the source voltage. Take a look at the following vector diagram.

Where the power factor of the circuit is

$\cos { \theta } =\frac{R}{Z}=\frac{12}{19.4}=0.619\quad lagging$

As from the above calculation, we have observed that inductive reactance is larger than capacitive, so the power factor is considered lagging.

$\cos{\theta}=0.619\\ \theta=\cos^{-1}{0.619}=51.8^{ o }\quad lagging$

## RLC Series circuit calculator :

## Conclusion:

- If resistor, inductor, and capacitor are connected in series AC circuit, the circuit will be called RLC series circuit.
- The phase difference between voltage and current adjusted by the difference between capacitive and inductive reactance.
- In impedance and voltage triangle, quantities are added vectorially.
- If inductive reactance is larger, the circuit will respond as if it is an inductive circuit and vice versa.