inductive and capacitive circuits -...

Inductive & Capacitive

Circuits

Subhasish Chandra Assistant Professor Department of Physics Institute of Forensic Science, Nagpur

… 2

LR CircuitLet I be the value of the current at any instant. The rate at which the current increases is dI/dt. The potential drop across the resistance is RI Back e.m.f introduced in the inductance is

Back e.m.f. opposes the e.m.f. E due to the battery. The net e.m.f is

Let us consider a circuit having an inductance L and a resistance R connected in a series with a battery of e.m.f E.

The current increases gradually from zero to a maximum value of I0. The time in which the current is increasing, a back e.m.f is introduced in the inductance.

−L dIdt

E − L dIdt

LR Circuit (Charging)www.electronics-tutorials.ws

Physics for Degree Students B.Sc. First Year by C.L. Arora & P. S. Hemne

… 3

LR CircuitHence, we get Applying Kirchhoff’s Voltage law, we

get

Integrating both sides, we get

where, K is the constant of integration

When t = 0, I = 0 and we get

E − L dIdt

= RI

⇒ E − L dIdt

= RI

⇒ ER− I = L

RdIdt

⇒ −dIER − I

= − RLdt

ln ER− I⎛

⎝⎜⎞⎠⎟ = − R

Lt + K

K = ln ER

⎛⎝⎜

⎞⎠⎟

ln ER− I⎛

⎝⎜⎞⎠⎟ = ln

ER

⎛⎝⎜

⎞⎠⎟ −

RLt

⇒ ln ER− I⎛

⎝⎜⎞⎠⎟ = ln

ER

⎛⎝⎜

⎞⎠⎟ −

RLt

⇒ lnER − IER

⎛

⎝⎜⎜

⎞

⎠⎟⎟= − R

Lt

⇒ER − IER

= e−RLt

⇒ ER− I = E

Re−RLt

⇒ I = ER1− e

−RLt⎛

⎝⎜⎞⎠⎟

⇒ I = I0 1− e−RLt⎛

⎝⎜⎞⎠⎟

I0 = E/R is the F i n a l S t e a d y Value of Current

∵dxx= ln x∫


… 4

LR CircuitThe fraction L/R is called the time constant of the circuit. If t = L/R = τ, then

Time constant of LR circuit is defined as the time during which the current rises by 63.21% of the maximum steady value.

We get At t = 0, I = 0 The current increases exponentially with time

As t → ∞ , t he cu r ren t t ends asymptotically to the steady state value I0 = E/R

I = I0 1− e−RLt⎛

⎝⎜⎞⎠⎟= I0 1− e

−RLLR

⎛⎝⎜

⎞⎠⎟

⇒ I = I0 1− e−1( )

⇒ I = 0.6321I0LR Circuit (I vs t) (Charging)

www.electronics-tutorials.ws


… 5

LR Circuit

After the steady current has been established, the steady e.m.f is removed. The current begins to decay exponentially and during this time a back e.m.f -L dI/dt is introduced in the inductance.

Applying Kirchhoff’s Voltage law, we get


where, K is the constant of integration When t = 0, I = I0 and we get

Hence, we get

−L dIdt

= RI

⇒ dII= − R

Ldt

ln I = − RLt + K

K = ln I0

ln I = ln I0 −RLt

⇒ ln II0

⎛⎝⎜

⎞⎠⎟= − R

Lt

LR Circuit (Discharging)www.electronics-tutorials.ws


… 6

LR CircuitFor t = L/R = τ,

Time constant of LR circuit can also be defined as the time during which the current decreases by 36.79% of the maximum steady value.

The current decays exponentially from its maximum value I0 to zero.

⇒ II0

= e−RLt

⇒ I = I0e−RLt

I = I0e−RLt= I0e

−RLLR

⇒ I = I0 e−1( )

⇒ I = 0.3679I0

LR Circuit (I vs t)(Discharging)www.electronics-tutorials.ws


… 7

LR CircuitThe reactance for inductance is XL = ωL = 2πfL The current is given as,

where,

Z is the impedance of the circuit and is given by 𝜙 is the phase difference between the current and voltage and is given by

The current lags with respect to the e.m.f.

Let us consider a circuit having an inductance L and a resistance R connected in a series with a power supply giving alternating e.m.f E.

Let the angular frequency of the alternating e.m.f. be ω Let the maximum magnitude of e.m.f be E0

The alternating e.m.f be E0 sin ωt

I = EZ= E0Zsin(ωt −φ)

I = E0R2 +ω 2L2

sin(ωt −φ)

LR Circuit (AC)

Z = R2 +ω 2L2

φ = tan−1 XL

R⎛⎝⎜

⎞⎠⎟ = tan

−1 ωLR

⎛⎝⎜

⎞⎠⎟



… 8

RC CircuitLet q be the value of the charge at any instant. The rate at which the current increases is dq/dt. The potential drop across the resistance is RI = R dq/dt The potential drop across the capacitance is q/C The capacitor plate charges to a constant maximum of q0, when the potential difference across the capacitor is E.


Let us consider a circuit having a capacitance C and a resistance R connected in a series with a battery of e.m.f E. The charge increases gradually from zero to a maximum value of q0.

R dqdt

+ qC

= E

⇒ R dqdt

+ qC

= q0C

∵E = q0C

RC Circuit (Charging)www.electronics-tutorials.ws


… 9

RC Circuit


where, K is the constant of integration When t = 0, q = 0 and we get

Hence, we get

− ln q0 − q( ) = tRC

+ K

⇒ dqq0 − q

= 1RC

dt

K = − ln q0( )

− ln q0 − q( ) = tRC

− ln q0( )

⇒ ln q0 − qq0

⎛⎝⎜

⎞⎠⎟= − t

RC

⇒ q0 − qq0

= e− tRC

⇒ q = q0 1− e− tRC

⎛⎝⎜

⎞⎠⎟ RC Circuit (I vs t) (Charging)



… 10

RC CircuitThe fraction RC is called the capacitative time constant of the circuit. If t = RC = τ, then

Time constant of RC circuit is defined as the time during which the charge rises to 63.21% of the maximum steady value.

The current is given as,

The steady current is given as

Hence, we observe that,

The charge in a RC circuit exponentially increases with time to a steady maximum value q0

The current decreases from a steady maximum value I0 to zero exponentially.

I = dqdt

= ddt

q0 1− e− tRC

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

⇒ I = q0RC

e− tRC

I0 =ER= q0RC

∴ I = I0e− tRC

q = q0 1− e− tRC

⎛⎝⎜

⎞⎠⎟= q0 1− e

−RCRC

⎛⎝⎜

⎞⎠⎟

⇒ q = q0 1− e−1( )

⇒ q = 0.6321q0


… 11

RC Circuit

After the steady current has been established, the steady e.m.f is removed. The charge in the capacitor begins to decay exponentially through the resistance and a current flows through the resistance.



where, K is the constant of integration When t = 0, q = q0 and we get

Hence, we get

lnq = − 1RC

t + K

K = lnq0

lnq = lnq0 −1RC

t

⇒ ln qq0

⎛⎝⎜

⎞⎠⎟= − 1

RCt

⇒ q = q0e− tRC

R dqdt

+ qC

= 0

⇒ dqq

= − dtRC

RC Circuit (Discharging)www.electronics-tutorials.ws


… 12

RC CircuitThe current is given as,

Hence, we observe that, The charge in a RC circuit exponentially discharges with time from a steady maximum value q0 to zero The current changes from a steady maximum value -I0 to zero exponentially.

For t = RC = τ,

Time constant of RC circuit can also be defined as the time during which the charge falls to 36.79% of the maximum steady value.

q = q0e− tRC = q0e

−RCRC = q0e

−1 = 0.3679q0

I = dqdt

= ddt

q0e− tRC

⎛⎝⎜

⎞⎠⎟

⇒ I = − q0RC

e− tRC = −I0e

− tRC

RC Circuit (I vs t) (Discharging)www.electronics-tutorials.ws


… 13

RC CircuitThe reactance for capacitance is XC = 1/ωC = 1/2πfC The current is given as,

where,

Z is the impedance of the circuit and is given by 𝜙 is the phase difference between the current and voltage and is given by

The current leads with respect to the e.m.f.

Let us consider a circuit having an capacitance C and a resistance R connected in a series with a power supply giving alternating e.m.f E.



RC Circuit (AC)

Z = R2 + 1/ω 2C 2( )

φ = tan−1 XC

R⎛⎝⎜

⎞⎠⎟ = tan

−1 1ωRC

⎛⎝⎜

⎞⎠⎟

I = EZ= E0Zsin(ωt +φ)

I = E0

R2 + 1ω 2C 2

sin(ωt +φ)www.electronics-tutorials.ws


… 14

LCR CircuitThe reactance for capacitance is XC = 1/ωC = 1/2πfC The reactance for inductance is XL = ωL = 2πfL The current is given as,

where, Z is the impedance of the circuit and is given by

Let us consider a circuit having an capacitance C, inductance L and a resistance R connected in a series with a power supply giving alternating e.m.f E.



LCR Circuit (Series - AC)

Z = R2 + ωL − 1ωC

⎛⎝⎜

⎞⎠⎟2


I = E0

R2 + ωL − 1ωC

⎛⎝⎜

⎞⎠⎟2sin(ωt −φ)



… 15

LCR CircuitHence, we get three scenarios,

XL > XC i.e.

The phase angle 𝜙 will be positive and the current lags behind the applied e.m.f.

The potential difference across the inductance is greater than that of the capacitor and the circuit will behave as an inductive circuit. XL < XC i.e.

The phase angle 𝜙 will be negative and the current leads the applied e.m.f. The potential difference across the inductance is lesser than that of the capacitor and the circuit will behave as an capacitive circuit.

𝜙 is the phase difference between the current and voltage and is given by

The phase difference depends on the relative values of XL and XC.

LCR Circuit (Series - AC)

φ = tan−1 XL − XC

R⎛⎝⎜

⎞⎠⎟ = tan

−1ωL − 1

ωCR

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

ωL > 1ωC

ωL < 1ωC



… 16

LCR CircuitThe resonant frequency f0 is given as,

XL = XC i.e.

The phase angle 𝜙 will be zero and the current is in phase with the applied e.m.f.

The potential difference across the inductance and capacitance is equal in magnitude but opposite in phase. Hence, the entire potential drops across the resistance and it behaves as a purely resistive circuit.

In this case the current is maximum i.e.

The frequency at which this occurs is called as the resonant frequency.

The circuit is said to be a series resonant circuit.

ωL = 1ωC

XL = XC ⇒ω 0L = 1ω 0C

⇒ω 0 =1LC

⇒ f0 =1

2π LC

I0 =E0R


XL < XC XL > XC


… 17

LCR CircuitThe reactance for capacitance is XC = 1/ωC = 1/2πfC The reactance for inductance is XL = ωL = 2πfL The current is given as,

where, Z is the impedance of the circuit and is given by

𝜙 is the phase difference between the current and voltage and is given by

Let us consider a circuit having an capacitance C connected in parallel with an inductance L and a resistance R. A power supply giving alternating e.m.f E is connected to this combination.



LCR Circuit (Parallel - AC) 1Z= jωC + 1

R + jωL


φ = tan−1I1XL

ZLR

− I2

I1RZLR

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= tan−1

I1ωL

R2 +ω 2L2− I2

I1R

R2 +ω 2L2

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟



… 18

LCR CircuitThe resonant frequency f0 is given as,

The current is minimum and impedance is maximum at resonance. Such a circuit rejects the currents at resonance and are known as rejector circuits, filter circuits or anti-resonant circuits.

For resonance, the current through the capacitance is equal to quadrature component of the current through the series combination of inductance and resistance. At this condition supply current I is in-phase with the supply voltage E. This condition can be represented as,

EXC

= EZLR

iXL

ZLR


XL < XC XL > XC

⇒ ZLR2 = XCXL

⇒ R2 +ω 02L2 = 1

ω 0Ciω 0L

⇒ R2 +ω 02L2 = L

C

⇒ R2 + 2π f02L2 =LC

f0 =12π

1LC

− R2

L2


inductive and capacitive circuits -...

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