# Inductors in series and parallel | Concepts you need to know and 10+ important problems

## Inductors

Inductors are nothing but magnetic energy storage devices. Physically it is a coil of conducting wire, either wrapped around a solid core or without any core. The latter one is called an air-core inductor.

When current flows through the inductor, it creates a magnetic field. Coiling up a lot of wire increases the strength of the magnetic field. The direction of the magnetic field is determined with the help of the right-hand thumb rule

When current first starts to flow through the coil, the magnetic field starts to expand, then after some time, it stabilizes and stores some amount of magnetic energy. When the field gradually collapses, the magnetic energy gets turned back into electrical energy. Inductors produce magnetic flux, proportional to the current flowing through them.

## Self inductance Definition

Self inductance is the characteristic of a coil by which the coil opposes any sudden change of current in it.

Self inductance of a coil, [Latex]L=\frac{N\phi }{i}[/Latex]

Where, N = number of turns in the coil, ? = magnetic flux and i is the current flowing through the coil

Self inductance of a solenoid with n turns, l length and A cross-sectional area, [Latex]L=\frac{N\phi }{i} = \frac{NBA}{i} = \frac{N}{i} \times \frac{\mu_{0}NAi}{l} = \frac{\mu_{0}N^{2}A}{l}[/Latex] ( Answer )

## Mutual Inductance Definition

In the case of two coils, the change in current in one coil induces EMF in the neighboring coil. This incident is known as mutual induction, and this property of the primary coil is called mutual inductance.

## Adding inductors in series | Two inductors in series

In a inductors in series connection, we can see from the diagram that the current in each inductor is equal. So the total voltage drop across the inductors is the sum of every individual inductor’s voltage drop. Suppose L is the total inductance of the circuit. So total voltage drop VTotal will be

VTotal = V1 + V2

The V1 and V2 is the voltage drop across the the individual inductor respectively.

By Kirchhoff’s rule we can write,

[Latex]V_{Total} -(L_{1}+L_{2})\frac{\mathrm{d} i}{\mathrm{d} t} =0[/Latex]

[Latex]V_{Total} =(L_{1}+L_{2})\frac{\mathrm{d} i}{\mathrm{d} t}[/Latex]

[Latex]L\frac{\mathrm{d} i}{\mathrm{d} t} =(L_{1}+L_{2})\frac{\mathrm{d} i}{\mathrm{d} t}[/Latex]

L=L1+L2

## The equivalent inductance of inductors in series  | Formula for inductor in series

Similar to the previously found equation for two inductors, if we connect n number of inductors in series with self inductance L1, L2, L3,…..Ln in series, the equivalent inductance for inductors in series circuit will be,

Leq = L1 + L2 + L3 + ….. + Ln

## Inductors in parallel

In a parallel connection, we can conclude from the diagram that the total current flowing through the circuit is the summation of the individual coil’s current. The voltage across each inductor is the same.

If the supply voltage is V then,

[Latex]V = L\left ( \frac{\mathrm{d} i_{1}}{\mathrm{d} t} + \frac{\mathrm{d} i_{2}}{\mathrm{d} t} \right )[/Latex]

[Latex]V = L\left ( \frac{V}{L_{1}} + \frac{V}{L_{2}}\right )[/Latex]

[Latex]\frac{1}{L}=\frac{1}{L_{1}} + \frac{1}{L_{2}}[/Latex]

[Latex]L = \left ( \frac{1}{L_{1}} + \frac{1}{L_{2}} \right )^{-1}[/Latex] ( Answer )

## The equivalent inductance of inductors in parallel  | Inductor in parallel formula

The equivalent inductance of n inductors with self inductance L1, L2, L3,…..Ln connected in parallel is,

[Latex]L_{eq} = \left ( \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} +…….\frac{1}{L_{N}}\right )^{-1}[/Latex] ( Answer )

## Inductors in series with mutual inductance

For the above derivations, we assumed that there is no mutual inductance in between the inductors. Now, if the inductors are connected in such a manner that the magnetic field produced by one affects the inductance of others, the inductors are said to be ‘mutually connected.’

## Coupled inductors in series

The magnetic fields of the inductors can be either aiding or opposing each other depending upon the orientation of the coils. Coupling can be classified into two types-

## Series Aiding Type of Coupling:

In this type of coupling, the magnetic fields of the inductors are in the same direction. So the currents that flow through the inductors are also in the same direction. For two inductors with self inductances L1 and L2 and mutual inductance M, we can write,

Total induced EMF = Self-Induced EMFs in L1 and L2 + induced EMF in one coil due to change of current in other for mutual inductance

[Latex]V = V_{1} + V_{2} + V_{M_{12}} + V_{M_{21}} = L_{1}\frac{\mathrm{d} i}{\mathrm{d} t} + L_{2}\frac{\mathrm{d} i}{\mathrm{d} t} + M\frac{\mathrm{d} i}{\mathrm{d} t} + M\frac{\mathrm{d} i}{\mathrm{d} t} = (L_{1} + L_{2} + 2M)\frac{\mathrm{d} i}{\mathrm{d} t}[/Latex]

Therefore,

The equivalent inductance =  L1+ L2 +2M

## Series opposing type of coupling:

In this type of coupling, the magnetic fields of the inductors are in the opposite direction. So the directions of the currents are opposite to each other. For two inductors with self inductances L1 and L2 and mutual inductance M, we can write,

Total induced EMF = Self-Induced EMFs in L1 and L2 + induced EMF in one coil due to change of current in other for mutual inductance

[Latex]V = V_{1} + V_{2} + V_{M_{12}} + V_{M_{21}} = L_{1}\frac{\mathrm{d} i}{\mathrm{d} t} + L_{2}\frac{\mathrm{d} i}{\mathrm{d} t} – M\frac{\mathrm{d} i}{\mathrm{d} t} – M\frac{\mathrm{d} i}{\mathrm{d} t} = (L_{1} + L_{2} – 2M)\frac{\mathrm{d} i}{\mathrm{d} t}[/Latex]

Therefore, equivalent inductance =  L1+ L2 -2M

## Impedance of capacitor and inductor in series LC circuit:

For the above capacitor and inductors in series circuit, we are going to assume that there is no resistance. We place a fully charged capacitor along with an inductor in the circuit. Initially, the switch is open. Suppose the capacitor plates have charge Q0 and -Q0

At t=0, the switch is closed. The capacitor begins to discharge , and the current starts increasing in the coils of the inductor with inductance L. Now, if we apply Kirchhoff’s law, we get,

[Latex]E + \frac{Q}{C} = 0[/Latex] (voltage drop across the inductor is E)

[Latex]-L\frac{\mathrm{d} i}{\mathrm{d} t} + \frac{Q}{C} = 0[/Latex]

[Latex]L\frac{\mathrm{d} ^{2}Q}{\mathrm{d} t^{2}} + \frac{Q}{C} = 0……(1)\,\, \: \left ( i=-\frac{\mathrm{d} Q}{\mathrm{d} t} \right )[/Latex]

A solution to this second order differential equation is,

[Latex]Q = Q_{0}cos(\omega t +\phi )[/Latex]  where Q0 and ? are constants depending on the initial conditions

If we put the value of Q in (1), we get,

[Latex]L\frac{\mathrm{d} ^{2}(Q_{0}\cos (\omega t +\phi)) }{\mathrm{d} t^{2}} + \frac{Q_{0}\cos (\omega t +\phi)}{C} = 0[/Latex]

[Latex]-L\omega ^{2}Q_{0}\cos (\omega t +\phi) + \frac{Q_{0}\cos (\omega t +\phi)}{C} = 0[/Latex]

[Latex]\left ( \frac{1}{C} – L\omega ^{2} \right )Q_{0}\cos (\omega t +\phi) =0[/Latex]

Therefore, [Latex]\frac{1}{C} – L\omega ^{2} =0[/Latex]

[Latex]\omega =\frac{1}{\sqrt{LC}}[/Latex]

[Latex]i= -\frac{\mathrm{d} Q}{\mathrm{d} t} = Q_{0}sin(\omega t +\phi )[/Latex]

## Energy stored in LC series circuit

For the above capacitor and inductors in series circuit

Total energy in LC circuit= energy stored in the electric field + energy stored in the magnetic field

[Latex]E = \frac{Q^{2}}{2C}+\frac{Li^{2}}{2}[/Latex]

[Latex]= \frac{(Q_{0}cos(\omega t+\phi ))^{2}}{2C}+\frac{L(Q_{0}\omega sin(\omega t+\phi ))^{2}}{2}[/Latex]

[Latex]= \frac{(Q_{0}cos(\omega t+\phi ))^{2}}{2C}+\frac{(Q_{0}sin(\omega t+\phi ))^{2}}{2C}[/Latex]      [since ⍵=1/LC ]

[Latex= \frac{Q_{0}^{2}}{2C}][/Latex]

## Impedance of capacitor and inductor in series | Impedance in LC circuit

For the above capacitor and inductors in series circuit

Total impedance of LC circuit XLC=XL-XC if XL>XC

=XC-XL if XL<XC

## An inductor and a capacitor are connected with a 120 V, 60 Hz AC source. For the following LC circuit, find total impedance and the current flowing through the circuit.

Given:

L= 300 mH    C = 50 µF    V = 120 V   f = 50 Hz

We know, XL= 2πfL and  XC= 1/2πfC

Putting the given value of L and C we get,

XL = 113 ohm

XC= 53 ohm

Therefore, total impedance, Z = XL – XC = 113 – 53= 60 ohm

Current in the circuit, i = V/Z = 120/60 = 2 A

1. An LC circuit consists of an inductor of L = 20mH and a capacitor of C = 50µF. The initial charge on the capacitor plate is 10mC. What is the total energy? Also, find out the resonance frequency.

Given:

L= 20 mH    C = 50 µF    Q0 = 10 mC

Total energy E = Q02/2C = (10 x .001)2 / 2x 0.00005 = 1 J

Resonance frequency f =1/2√LC= 1/(2 x 3.14 x √(20 x 0.001 x 0.00005)) = 159 Hz ( Answer )

## Resistor and inductor in series LR circuit

Circuits containing resistors and inductors are known as LR circuits. When we connect a voltage source, the current starts flowing through the circuit. Now, if we apply Kirchhoff’s law, we get,

[Latex]V_{0}-iR – L\frac{\mathrm{d} i}{\mathrm{d} t}=0[/Latex]   ( V0 is the voltage of the source)

[Latex]V_{0}=iR + L\frac{\mathrm{d} i}{\mathrm{d} t}[/Latex]

[Latex]\frac{di}{V_{0}-iR}=\frac{dt}{L}[/Latex]

Integrating both the sides with limit i = 0 to I and t = 0 to t , we get,

[Latex]\frac{-\ln (V_{0}-iR) + \ln (V_{0})}{R}=\frac{t}{L}[/Latex]

[Latex]\ln (\frac{V_{0}-iR}{V_{0}})=\frac{-Rt}{L}[/Latex]

[Latex]\frac{V_{0}-iR}{V_{0}}=e^{\frac{-Rt}{L}}[/Latex]

## Time constant of LR circuit

? = L/R is called the time constant of LR circuit

## Impedance of inductor and resistor in series | Impedance of LR circuit

The resistance and the inductance are the components responsible for the total impedance of the LR circuit.

The total impedance, [Latex]Z=\sqrt{R^{2}+X_{L}^{2}}[/Latex] ( Answer )

## A 24 V battery is removed from a circuit consisting of a resistor with 2-ohm resistance and an inductor with 0.03 H inductance. Calculate the initial current at t = 0 second. Find out how long it takes for the current to decrease to 50% of the initial current.

If the battery is suddenly removed from the circuit, then the current takes some time before dropping to zero.

At t = 0, i = V0/R = 24/2 = 12 A

Time constant ? = L/R = 0.03/2 = 0.015 second

i = i0e-t/? where I0 is the initial current before closing the switch

0.5 = e-t/0.015

t/0.015 = -ln(0.5)

t = 0.01 s ( Answer )

## A 2 Ohm resistor and an 8 mH inductor are connected in series with a power supply of 6 volts. How much time will it take for the current to become 99.9% of the final current?

Time constant of the circuit = L/R = 8 x 0.001 / 2 = 4 ms

I = Ifinal x 99.9/100

Ifinal (1 – e-t/?) = Ifinal x 0.999

1 – e-t/? = 0.999

e-t/? = 0.001

t/? = 6.9

t= 6.9 x 4 = 27.6 ms ( Answer )

## The impedance of resistor, capacitor, and inductor in series RLC circuit

The above has a resistor, an inductor, and a capacitor connected in series with an AC source. When the circuit is in the closed condition, the electric current starts to oscillate sinusoidally. This phenomenon is analogous to the spring-mass system in simple harmonic motion.

If we apply Kirchhoff’s law in the circuit, we get,

[Latex]\frac{Q}{C}-L\frac{\mathrm{d} i}{\mathrm{d} t}-iR=0[/Latex]

[Latex]\frac{Q}{C}+L\frac{\mathrm{d}^{2} Q}{\mathrm{d} t^{2}}+R\frac{\mathrm{d} Q}{\mathrm{d} t}=0\: \: \, \: \: \; (i=-\frac{\mathrm{d} Q}{\mathrm{d} t})[/Latex]

[Latex]L\frac{\mathrm{d}^{2} Q}{\mathrm{d} t^{2}}+R\frac{\mathrm{d} Q}{\mathrm{d} t}+\frac{Q}{C}=0[/Latex]

Now, comparing this with the equation of damped harmonic motion, we can get a solution here.

[Latex]Q=Q_{0}e^{\frac{-Rt}{2L}}cos(\omega ‘t+\phi )[/Latex]

## Impedance of a series RLC circuit

A RLC circuit has three elements responsible for total impedance.

1. Resistor impedance R
2. Capacitor impedance or capacitive reactance XC = 1/⍵C = 1/2πfC
3. Inductor impedance or inductive reactance XL = ⍵L = 2πfL

Therefore, total impedance, [Latex]Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}[/Latex] ( Answer )

## A series RLC circuit consists of a resistor of 30 ohm, an inductor of 80 mH and a capacitor of 40 µF. It is given an AC supply voltage of 120 V and 50 Hz. Find out the current in the circuit.

Solution :

Inductive reactance XL= 2πfL = 2 x 3.14 x 80 x 0.001 x 50 = 25.13 ohm

Capacitive reactance XC = 1/2πfC = 79.58 ohm

Total impedance, Z = √{R2 +(XC – XL)2}= √{(30)2 +(79.58-25.13)2} = 62.17 ohm

Therefore, current in the circuit, i = 120/62.17 = 1.93 A

1. Derive the equation for current in the below circuit where V= sin4t

Applying Kirchhoff’s law in the circuit, we can write,

Sin4t – 3i – 2di/dt + Q/0.5 = 0

Sin4t = 3i + 2di/dt + 2Q

Taking differentiation on both the sides,

4cos4t = 3di/dt + 2d2i/dt2 +2 i(t)

i(t) + 3/2(di/dt) + d2i/dt2 = 2cos4t       This is the required equation for current. ( Answer )

## 1. An LC circuit stores a total energy of E. Maximum charge on the capacitor is Q . Energy stored in the inductor while the charge on the capacitor is Q/2 is

1. E
2. E/2
3. E/4

Solution:  Total energy = E = Q2/2C

Total energy = EC + E

When, charge on the capacitor is Q/2, total energy,

Q2/2C  =  (Q/2)2/2C + Ei

Ei = Q2/2C x (1-¼) = 3E/4    ( Answer )

## 2. If the current in one coil becomes steady, what would be the current flowing through the neighboring coil?

1. Double of first coil
2. Half of first coil
4. Infinity

Solution: Current is induced when magnetic flux in coil changes. Hence, if the current is steady in one coil, no flux will be generated and current in the neighboring coil will be zero.

## 3. A 7 ohm resistor is connected in series with a 32 mH inductor in inductors in series circuit. If the supply voltage is 100 volt, 50 Hz then calculate the voltage drop across the inductor.

1. 67 V
3. 54 V
4. 100 V

Detailed Solution of the problem:

The inductive reactance XL for the circuit = 2 x 3.14 x 50 x 0.032 = 10 ohm

Total impedance Z = (R2 + XL2) = (72 + 102) = 12.2 ohm

Therefore, current across the circuit = 100/12.2 = 8.2 A

The voltage drop across the inductor = iXL = 8.2 x 10 = 82 V  (Answer)

## 4. Find the equivalent impedance for the infinite ladder circuit shown below-

1. j4 ohm
2. j8 ohm
3. j4(√2 – 1) ohm
4. j4(√2 + 1) ohm (Answer)

Solution:  For the above infinite circuit let us assume that,

Z1 = j8 ohm and Z2 = j4 – j2 = j2 ohm

If the equivalent impedance is Z then, we can write

Z = Z1 + (Z2 || Z) = Z1 + ZZ2/Z + Z2

Z( Z + Z2 ) = Z1Z2 + ZZ1 + ZZ2

Z2 + j2Z = -16 + j8Z + j2Z

Z2 – j8Z + 16 = 0

Solving the quadratic equation, we get,

Z = j4(√2 + 1) ohm (Answer)

## 5. Self inductance of a solenoid is 5 mH. The coil has 10 turns. What will be the inductance of the coil if the number of turns is doubled?

1. 10 mH
2. 5 mH
4. 30 mH

Solution: Self inductance of the solenoid with N turns and A cross-sectional area is = μ0N2A / l

Here μ0 x 100 x A / l = 5

μ0A / l = 1/20

If the number of turns is doubled then new self inductance = μ0A / l x N’2 = 1/20 x (20)2 = 20 mH (Answer)

## How to add inductors in series and parallel? | Inductors in series vs parallel:

In series, the sum of the self inductance of all the inductors is the total inductance of the circuit. For parallel connection, the sum of the inverse of all the self inductances is the inverse of the total inductance.

## How does adding inductors in series to a circuit affect the current?

Inductors added in the series share the same current. Thus the total voltage of the circuit is higher than the voltages of individual inductors.

## What are differentially coupled series inductors?

It is another name for the series opposing inductors where the magnetic fluxes produced by the inductors are opposite in direction. The total inductance is in this type of inductor is the sum of self inductance of the inductors – 2 x the mutual inductance.

## What is the mutual inductance of two coils in series?

Mutual inductance of two iron-core coils with turns N1 and N2, cross-sectional area A, length L and permeability μr is, [Latex]M = \frac{\mu {0}\mu {r}N_{1}N_{2}A}{L}[/Latex]

## What is series inductor filter?

Series inductor filter is an inductor connected in series in between the load and the rectifier. It is called a filter as it blocks AC and allows DC.

## An inductor of 1 henry is in series with a capacitor of 1 microfarad. Find the impedance when the frequency is 50 Hz and 1000 Hz.

Impedance, Z = XL – X

XC when frequency is 50 Hz = 1/2πf1C = 3183 ohm

XC when frequency is 1000 Hz = 1/2πf2C = 159 ohm

XL when frequency is 50 Hz = 2πf1L = 314 ohm

XL when frequency is 1000 Hz = 2πf1L = 6283 ohm

Therefore, impedance Z1 when frequency is 50 Hz = 6283 – 159 = 6124 ohm

impedance Z2 when frequency is 1000 Hz = | 314 – 3183 | = 2869 ohm.

Kaushikee Banerjee

I am an electronics enthusiast and currently devoted towards the field of Electronics and Communications . My interest lies in exploring the cutting edge technologies. I'm an enthusiastic learner and I tinker around with open-source electronics. LinkedIn ID- https://www.linkedin.com/in/kaushikee-banerjee-538321175