Worked examples

WORKED EXAMPLES

Example 1 Two coupled coils with L₁ = 0.02 H, L₂ = 0.01 H and K = 0.5 are connected in four different ways, series aiding series opposing and parallel with both arrangements of the winding sense. What are the four equivalent inductances?

Solution:

M = K√L₁ L₂

= 0.5 √0.02 × 0.01

= 7.07 × 10^-3 H = 0.00707 H

 (a) For a series aiding

Total inductance = L₁+ L₂ + 2M

= 0.02 + 0.01 + 2 (0.00707)

= 0.044 H

(b) For a series opposition

Total inductance = L₁ L₂ – M² / L₁+ L₂ – 2M

= 0.02 + 0.01-2(0.00707) = 0.016 H

(c) For parallel aiding

Total inductance

L₁ L₂ – M² / L₁+ L₂ – 2M

= 0.02 × 0.01 - (0.00707)² / 0.016

M = 1.5 × 10^-4 / 0.016 = 9.4 mH (milli Henrys)

(d) For parallel opposition

Total inductance =  L₁ L₂ – M² / L₁+ L₂ + 2M

= 1.5 × 10^-4 / 0.044

= 3.41 milli Henrys

Example 2(a) What is the maximum possible mutual inductance of two inductively coupled coils with self inductances

L₁ L₂ = 25 mH

L₂ = 100 mH ?

Solution:

M_max = √ L₁ L₂ since K = 1 for maximum M

√25 × 100 = 50 milli Henrys

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Example 2(b) Two inductors are connected as shown in fig. 5.34. What is the value of the effective inductance of the combination?

Solution: Effective inductive L = L₁ + L₂ – 2M

= 3 + 5 – 2 × 2

= 4 H

 [Note: As the current enters the first coil at dotted terminal and leaves the other coil as dotted terminal, M is taken as negative.]

Example 3 Two coupled coils with respective self inductances L₁ = 0.8 H and L₂ = 0.2H have a coupling coefficient of 0.6. Coil 2 has 500 turns. If the current in coil 1 is i1 (t) = 10 sin 200 t, determine the voltage at coil 2 and the maximum flux set up by the coil 1.

Solution:

Let the flux in coil 2, due to a flux ϕ₁ in coil 1 be ϕ₁₂. Then, e₂ is also obtained in the following

= 0.96 /200 sin 200 t webers

= 4.8 × 10^-3 x sin 200 t webers

ϕ₁₂ is maximum when sin 200 t is unity.

The maximum value of the flux ϕ₁₂ = 4.8 × 10^-3 webers

By definition, K = ϕ₁₂ / ϕ₁

= maximum value of ϕ₁₂ / maximum value of ϕ₁

maximum value of ϕ₁ = maximum value of ϕ₁₂ / K

= 4.8 × 10^-3 / 0.5.

= 9.6 × 10^-3 webers

= 9.6 milli webers

Example 4 Show that the circuit shown in fig. 5.35 behaves with respect to terminals A and B, as an open circuit or as a short circuit if

Solution: With respect to terminals A and B, the network behaves as open circuit if the net reactance between A and B (X_AB) is infinite. The network behaves like a short circuit if X Hence, first let us find out X_AB. The conductively coupled circuit of the given network is shown in fig. 5.36

Hence, when X_c = X₂, the network with respect to terminals A and B behaves like open circuit.

i.e., for this value of X, the network with respect to terminals A and B behaves like a short circuit.

 Example 5 In the coupled circuit shown in Fig. 5.37, find the voltage across the 5 ohm resistor.

Solution:

Conductivity coupled equivalent circuit is drawn as below for convenience.

Now by inspection,

Example 6 In the circuit find the phasor voltage V₂

Solution: Here mesh equations are written directly,