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Circuit theory is the linear analysis of voltage and current relationships, in any electrical circuit. There are applications of the concepts of inductance, capacitance, and resistance, in circuit theory. The role of these circuit elements in an electrical circuit and how they influence the current, voltage relationship will be studied under this topic. Let us talk about the purely resistive and purely inductive circuits.

Purely Resistive Circuit

The potential difference in a circuit, V = V0Sinωt…(1)

Current flowing through a purely resistive circuit is given by

i = i0Sinωt….(2)

Since the phase difference between the current and potential is zero so V ∝ I.

Purely Inductive Circuit

We know that in the inductor, the current lacks the potential by 90°.

So, i0 = \[\frac{ω}{Xl}\]= 2πfL

Here, we will study the combined Resistance and Inductor circuit, i.e., RL circuit.

RL Circuit is a Resistor – Inductor circuit, or RL filter, or RL network. It is an electric circuit that comprises passive components of resistors and inductors driven by a voltage or current source.

Now, let us understand the various types of RL circuits, i.e., RL series circuits.

[Image to be added soon]

In this circuit, the resistor and the inductor are connected in series.

Let us say current ‘i’ flows through the circuit, so the potential difference across the resistor and the inductor is VR and VL, respectively.

So, the potential across the resistor is:

VR(t) = (V0)R Sinωt

The potential across the inductor is:

VL(t) = (V0)L Sin (ωt + \[\frac{\pi}{2}\] ) (∵ Voltage leads current by \[\frac{\pi}{2}\])

So, the resultant potential is calculated by using the following formula:

Vo = \[\sqrt{(V_{o})r^{2}+(V_{o})l^{2}}\]

= \[\sqrt{(IR^{2})+(IoXl^{2})}\]

Vo = Io \[\sqrt{R^{2}+Xl^{2}}\]

where, Z = \[\sqrt{R^{2}+Xl^{2}}\]

So,

Z = Impedance of the RL circuit

So, the impedance of the RL circuit is the measure of the obstruction that a circuit presents (provides) to a current when a voltage is applied.

Here, we can see that the overall current still lacks, now let us understand the reason behind it?

If we look at the phasor diagram below

[Image to be added soon]

We know that the current lacks the potential across the inductor by 90°, but the resultant voltage, i.e., the resistance across the potential decreases the phase difference between the current and potential difference, VL by Φ. This happens because the resistive circuit tries to bring the phase difference between the current and the potential to zero.

Now, in the inductive circuit, the current lacks the potential by Φ instead of 90°.

Now, looking at the graph between the resistance and inductance, the resultant between XL (or the inductive reactance) and R is the impedance.

[Image to be added soon]

Now, let us derive the relationship between the inductance and the resistance by using the phasor diagram.

[Image to be added soon]

Now, finding the phase difference, we have:

An LR circuit is a serial arrangement of an inductor and a resistor. This circuit is connected with a voltage source, i.e., a battery and a switch. Here, the wire of the coil of an inductor has a DC resistance, i.e., R.

Let us take a look at the LR series circuit drawn below.

[Image to be added soon]

At t = 0, the circuit is open, the inductor behaves like an open circuit.

After switching on the circuit, it gets a step response type input voltage. The current starts flowing through the circuit but does not reach its peak value (as determined by the Ohm’s law (V/R)).

This current does not increase suddenly because of the self-induced emf in the coil that opposes its increase. This emf is generated because of the varying or increasing magnetic flux across the coils of an inductor. After some time, the voltage source neutralizes the effect of self-induced emf because of which the current flow becomes constant and the effect of induced current and the magnetic field reaches zero.

We can use Kirchhoff’s Voltage Law or KVL to define the voltage drop that persists in the circuit. The expression is as follows:

V(t) - (VR + VL) = 0

The voltage drop by using the Ohm’s law is I * R,

So, VR = I * R, and,

The voltage drop across the resistor is given as:

VL = L \[\frac{di}{dt}\]

The final expression of the voltage drop around the LR circuit is:

V(t) = I * R + L \[\frac{di}{dt}\]….eq (a)

We can see that the voltage drop across the resistor depends on current ‘i’, while the voltage drop across the inductor depends on the rate of change of current didt.

At t = 0, the current ‘i’ is zero, the above expression (equation (a)), which is also a first-order differential equation, can be rewritten to give the value of the current at an instant as:

Where,

V = Voltage in Volts

I = Current in Amperes

L = Inductance in Henries

R = Resistance in Ohms

t = Time in seconds

e = Base of natural logarithm whose value = 2.17828

The time constant, て= R/L is the Helmholtz equation governing the growth of current in the LR circuit.

FAQ (Frequently Asked Questions)

Q1: Where do We Use the RL Circuit?

Ans: An RL circuit is a series combination of the resistor and inductor.

We use this circuit in the chokes of luminescent tubes.

For supplying DC power to Radio Frequency amplifiers.

Q2: Write the Applications of the Inductor.

Ans: We find the application of inductor in the following ways:

The inductor helps in reducing the input voltage in an A.C. circuit without the loss of energy.

Under DC conditions, it serves as a static resistance in an A.C. circuit.

Q3: What is the RL Time Constant?

Ans: In simple terms, RL time constant in seconds is the ratio of the inductance in Henries to the resistance of the circuit in Ohms. It depicts the amount of time that it takes to conduct 63.2% of the current resulting in applying a voltage across an inductor.

Q4: A Current of 5 A is increasing at the Rate of 6 A/s through a Coil of Inductance 3 H. The Energy Stored in the Inductor per unit Time will be?

Solution: Here, I = 5 A, di/dt= 6 A/s, L = 3 H, dU/dt= ?

Energy stored by an inductor in a unit time is dU/dt, and

U = 1/2LI^{2}

So, dU/dt= d/dt(1/2LI^{2})

= LI di/dt….(1)

Putting the given values in equation (1), we get,

dU/dt= 4 * 5 * 3 = 60 W