IEMDaily - Video Lecture Notes

Example

Equivalent Inductance of infinite inductors in Parallel

By applying KCL in the above circuit we obtain

i = i 1 + i 2 + i 3 + . . . . . + i n

And since

i k = \frac{1}{L k} \int_{0}^{t} ν d t + i k (0)

we have

i = \frac{1}{L 1} \int_{0}^{t} v d t + i l (0) + \frac{1}{L 2} \int_{0}^{t} v d t + i 2 (o) + \frac{1}{L 3} \int_{0}^{t} v d t + i 3 (0) + . . . . . . . . + \frac{1}{L n} \int_{0}^{t} v d t + i n (0)

= (\underset{\frac{1}{L e q}}{\underset{⏟}{\frac{1}{L 1} + \frac{1}{L 2} + \frac{1}{L 3} + . . . . . . . + \frac{1}{L n}}}) \int_{0}^{t} v d t + \underset{i (0)}{\underset{⏟}{i 1 (0) + i 2 (0) + i 3 (0) + . . . . + i n (0)}}

= \frac{1}{L e q} \int_{0}^{t} v d t + i (0)

L_{e q}

in the above expression is the value of the equivalent inductance.

Definition

Star-delta transformation of inductors

The star and delta circuits shown in the figures are interconvertible.
The equivalent value of inductance for star to delta transformation is:

L_{R S} = \frac{L_{R} L_{S} + L_{S} L_{T} + L_{R} L_{T}}{L_{T}}

L_{R T} = \frac{L_{R} L_{S} + L_{S} L_{T} + L_{R} L_{T}}{L_{S}}

L_{S T} = \frac{L_{R} L_{S} + L_{S} L_{T} + L_{R} L_{T}}{L_{R}}

The equivalent value of inductance for delta to star transformation is:

L_{R} = \frac{L_{R S} L_{R T}}{L_{R S} + L_{R T} + L_{S T}}

L_{S} = \frac{L_{R S} L_{S T}}{L_{R S} + L_{R T} + L_{S T}}

L_{T} = \frac{L_{S T} L_{R T}}{L_{R S} + L_{R T} + L_{S T}}

Definition

Non-ideal inductor

The two contributions to the non-ideal behaviour of inductors are as follows:

The finite resistance of the wire which is used to wind the coil.
The cross-turn effects which become important at high frequencies.

Example

Behaviour of RL circuits at the time of switching

Example:
A solenoid of inductance

L

with resistance

r

is connected in parallel to a resistance

R

. A battery of emf

E

and of negligible internal resistance is connected across the parallel combination as shown in the figure. At the time

t = 0

, switch

S

is opened, calculate current through the solenoid after the switch is opened.Solution:
Initially, inductor is fully charged and acts as short circuit. Hence, initially current in the inductor is given by:

i (0_{-}) = \frac{E}{r}

When the switch is opened, current across the inductor does not change suddenly.
Hence, current just after opening the switch through the solenoid is given by:

i (0_{+}) = \frac{E}{r}

Example

Behaviour of RL circuit in steady state

Example:
A coil of resistance R and inductance L is connected to a battery of e.m.f. E volt. Find the final current in the coil.
Solution:
In steady state, inductor will carry maximum current and it behaves as a short circuit. Hence, the final current is given by: