An LCR circuit is also a resonant or tuned circuit. It consists of an inductor-L, capacitor-C, and resistor-R connected in either series or parallel. In this article, we will learn about an electrical circuit where an AC voltage is applied across an inductor, a capacitor, and a resistor connected in series. Table of Content Derivation of AC Voltage Applied to Series LCR Circuit Frequently Asked Questions-FAQs Derivation of AC Voltage Applied to Series LCR Circuit Consider the circuit shown above. Here, we have an inductor, a resistor, and a capacitor connected through a series connection across an AC voltage source given by V. Here, the voltage is sinusoidal in nature and is given by the equation, LCR series circuit AC voltage applied across an LCR circuit Here, vm is the amplitude of the voltage, and ω is the frequency. If q is the charge on the capacitor and i the current at time t, we have, from Kirchhoff’s loop rule: LCR series circuit Here, q is the charge held by the capacitor, I is the current passing through the circuit, R is the resistance of the resistor and C is the capacitance of the capacitor. To determine the instantaneous current or the phase of the relationship, we will follow the analytical analysis of the circuit. Analytical solution As i LCR series circuit , we can write LCR series circuit Hence, by writing the voltage equation in terms of the charge q through the circuit, we can write, LCR series circuit The above equation can be considered analogous to the equation of a forced, damped oscillator. To solve the equation, we assume a solution given by, LCR series circuit So, LCR series circuit And LCR series circuit Substituting these values in the voltage equation, we can write, LCR series circuit Here, we have substituted the value of Xc and XL by Xc = 1/ωC and XL = ω L. As we know, LCR series circuit hence, substituting this value in the above equation, we get, LCR series circuit Now, let LCR series circuit So we can say, LCR series circuit Now, comparing the two sides of the equation, we can write, LCR series circuit And, LCR series circuit Hence, the equation for current in the circuit can be given as, LCR series circuit Also read: Impedance of an LCR Circuit - GMS

We know that a capacitor is a passive electronic device with two terminals. It has the capacity to store electrical energy in an electrical field. In a DC circuit, when a capacitor is connected to a voltage source, the current will flow for the short time required to charge the capacitor. In this section, we will learn the expression of the AC voltage source applied across a capacitor in detail.

Table of Contents:

AC Voltage Source Applied Across a Capacitor

Let us consider the electric circuit shown below. We have a capacitor and an AC voltage V, represented by the symbol ~, that produces a potential difference across its terminals that varies sinusoidally. Here, the potential difference or the AC voltage can be given as,

\begin{array}{l} v = v_{m} s i n ω t \end{array}

Here, vm is the amplitude of the oscillating potential difference and the angular frequency is given by ω. The current through the resistor due to the present voltage source can be calculated using the Kirchhoff’s loop rule, as under,

\begin{array}{l} \sum V (t) = 0 \end{array}

For the given capacitor we can write,

\begin{array}{l} v = \frac{q}{C} \end{array}

According towe can write from the above circuit,

\begin{array}{l} v_{m} s i n ω t = \frac{q}{C} \end{array}

The current through the circuit can be calculated using the relation,

\begin{array}{l} i = \frac{d q}{d t} \\ \Rightarrow i = \frac{d (v_{m} C s i n ω t)}{d t} = ω C v_{m} c o s ω t \\ \Rightarrow i = i_{m} s i n (ω t + \frac{π}{2}) [U s i n g t h e r e l a t i o n, c o s ω t = s i n (ω t + \frac{π}{2})] \end{array}

Here the amplitude of the current can be written as,

\begin{array}{l} i_{m} = ω C v_{m} \end{array}

or else we can write it as,

\begin{array}{l} i_{m} = \frac{v_{m}}{\frac{1}{ω_{C}}} \end{array}

Here, we can see that the term 1/ωC can be said to be equivalent to the resistance of this device and is termed as the capacitive reactance. We denote the capacitive reactance of the device as XC.

\begin{array}{l} X_{C} = \frac{1}{ω_{C}} \end{array}

And thus, we can say that the amplitude of the current in this circuit is given as,

\begin{array}{l} i_{m} = \frac{v_{m}}{X_{C}} \end{array}

In the above equations, the dimension of the capacitive In the above equations, the dimension of the capacitive reactance can be seen to be the same as that of resistance, and also, the SI unit of capacitive reactance is given as ohm. The capacitive reactance restricts the passage of current in a purely capacitive circuit in the same way as resistance hinders the passage of current in a purely resistive circuit.

Here we say, that the capacitive reactance is inversely proportional to the frequency and the capacitance. We also see from the above equations that the current in a capacitive circuit is π/2 ahead of the voltage across the capacitor.

The instantaneous power supplied to the capacitor can be given in terms of the current passing through the capacitor as,

\begin{array}{l} P_{c} = i v = i_{m} c o s ω t v_{m} s i n ω t \\ P_{c} = \frac{i_{m} v_{m}}{2} s i n 2 ω t \end{array}

Here, the average power supplied over a complete cycle can be given as,

\begin{array}{l} P = \frac{i_{m} v_{m}}{2} s i n 2 ω t = 0 \end{array}

Frequently Asked Questions – FAQs

What is AC current?

An electric current that reverses direction periodically and changes its magnitude continuously with respect to time is known as an alternating current (AC).

What is the SI unit of capacitance?

The SI unit of capacitance is Farad.

State Kirchhoff’s Voltage Law.

Kirchhoff’s Voltage Law states that “the voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero”.

State true or false: The capacitive resistance is inversely proportional to the frequency and the capacitance.

True.

What is the expression for capacitive reactance?

The capacitive reactance of a circuit is given by,
XC =1/ωC
Where, ω is the angular frequency of the AC source
C is the capacitance of the capacitor

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