EFFECTS OF SELF INDUCTION A DC CIRCUIT

EFFECTS OF SELF INDUCTION IN A DC CIRCUIT

In case of non-inductive coil, when a pd of V volts is applied across it, current immediately attains the value of V/R amperes, when R is the resistance of the coil, because magnetic field is not set up due ti flow of current in such a coil, so there is no self induced emf which may oppose the applied voltage V.

But in case of an inductive coil, when potential difference  of V volts is applied across it, current starts rising, but due to rise in current flowing through the coil, self induced emf is produced in the coil, which applied voltage and increase in current. Theoretically the current takes infinite time to reach its maximum value given by .

 I_{max}=\dfrac{V}{R}

In case of non-inductive coil applied voltage is to overcome the resistive drop, IR only but in case of inductive coil applied voltage is to overcome the back emf set up by self induction of coil in addition to resistive drop IR.

Rise of Current

Consider an inductive coil of resistance R ohms and inductance L henrys connected across a source of supply of V volts.

Rise of current in inductor
fig: Rise of current

Let at any instance current flowing through the coil be of i amperes.

Resistive drop = i R volts

Back emf,  e'= -L\dfrac{di}{dt}

Applied voltage is to overcome resistive drop as well as back emf. Hence while current is increasing, we have

 V=i R+L\dfrac{di}{dt}

or  V-i R=L\dfrac{di}{dt}

or \dfrac{di}{{V-i R}} =\dfrac{dt}{L}

or \dfrac{-R di}{V-iR} =\dfrac{-R}{L}dt

Integrating both sides we get

log_e(V-i R)=\dfrac{-R}{L}t+K

where K is a constant of integration

From initial conditions, when t=0, i=0

 \therefore K=log_e V

Hence  \ log_e (V-i R)=\dfrac{-R}{L}t\log_e V

or  log_e \dfrac{V-iR}{V}=\dfrac{-R}{L}t

or \dfrac{V-iR}{V}= e^{\dfrac{-R}{L}t}

 V-i R=V.e^{\dfrac{-R}{L}t}

or iR=V\left( I-e^{\dfrac{-R}{L}t} \right)

or i=\dfrac{V}{R}\left( 1-e^{\dfrac{-R}{L}t}\right)=I_{max}(1-e^{-t/\tau}) … (Eqn 1)

rise and decay of current in inductor

where = I_{max}\dfrac{V}{R}, the final steady-state value of current and \tau=\dfrac{L}{R}, the time constant

(The time constant is represent is also represented by English letter “T”)

The equation  i=I_{max}(1-e^{-t/\tau}) is an exponential expression and is often referred to as Helmholtz express.

Immediately after the switch is closed, the rate of rise of current i.e., initial rate of rise of current can be determined by differentiating Eq.(1) w.r.t for t=0. Hence we get

\dfrac{di}{dt}=\dfrac{V}{R\tau}e^{t/\tau}=\dfrac{V}{R \times \frac{L}{R}}e^0=\dfrac{V}{L}

\therefore\tau=\dfrac{L}{R}and t=0

If this rate of rise of current were maintained then time required to attain the maximum value of current I_{max}=V/R word have been \dfrac{V/R}{V/L}=\dfrac{L}{R}. This time is called the time constant of the circuit.

Hence the time constant, \tau may be define as the time required for the current to rise to its final steady-state value if it continued rising at its initial rate (i.e.V/L)

If we substitute  t=\tau=\dfrac{L}{R} in Eq.(1),we get

i=I_{max}(1-e^{-1})=0.632 I_{max}

So the time constant may also be defined as the time required for the current to reach 0.632 of its final steady value (i.e.0.632 I_{max} ).

Decay of Current

Consider an inductive coil of resistance R ohms, inductance L henrys and having a current of I amperes flowing through. If the source of supply is re moved from its circuit, the current starts decreasing and due to decrease in current, self induced emf will be set up, which will oppose the decay of current.

Let at any instant current  following through the coil be of i  amperes

Self induced emf, e'=-L\dfrac{di}{dt}

Resistive drop=i R

since Applied voltage V=0

Since applied voltage,V=Resistive drop+back emf

0=i R+L\dfrac{di}{dt}

or \dfrac{di}{i}=\dfrac{-R}{L}dt

Integrating both sides we get

 \log_e i=\dfrac{-R}{L}t+K'

where K’ is a constant of integration

From initial conditions i.e. when t=0,  i=I_{max}

\therefore k'=\log_e I_{max}

Hence \log_e i=\dfrac{-R}{L}t+\log_e I_{max}

or \log_e\dfrac{i}{I_{max}}=\dfrac{-R}{L}t

or  i=I_{max}e^{\dfrac{-R}{L}t}=I_{max}e^{-t/ \tau}

where  \tau =\dfrac{L}{R}, the time constant.