Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion

1st case. Open circuit 3&4.

Since terminals of the secondary are open, no current flows through the secondary.

Hence,
V1 = Lp di1/dt,
Usually, V1 = Leq di1/dt,
From the qn, Leq = 4.
Comparing the above two, we get Lp = 4.

2nd case short circuit 3&4. Current flows out of the coil and V2 = 0,
V1 = Lp di1/dt - Mdi2/dt ---> eq2.

Lsdi2/dt - Mdi1/dt,
di2/dt = (M/Ls)di1/dt ---> eq 3.
Put eq 3 in eq 2.

V1 = Lp di1/dt - M(M/Ls)di1/dt,
V1 = ( Lp - (M*M /Ls)) di1/dt ---> eq 4.
But V1 = Leq di1/dt.

Here Leq =3(from qn).
Therefore, 3 = Lp - (M*M)/Ls.
Lp = 4(deducted earlier) and Ls = 2 ( from qn),
On substitution,
3 = 4 -(M*M)/2,
6 = 8-(M*M).
M*M = 2.

From the eqn of coupling coefficient,
K^2 = M^2 / (L1*L2),
K^2 = 2/ (4*2),
K^2 = 1/4,
K = 1/2 .
K= 0.5 (Answer).

1st case. Open circuit 3&4.
Kvl in loop1.
V1 = Lp di1/dt ---> eq1.

2nd case short circuit 3&4.
Kvl in loop1.
V1 = Lp' di2/dt +Mdi1/dt ---> eq2

Kvl in loop 2

Lsdi2/dt - Mdi1/dt
di2/dt = (M/Ls)di1/dt ---> eq 3
Put eq 3 in eq 2

V1 = Lp' di1/dt + M(M/Ls)di1/dt
V1 = ( Lp' + (M*M /Ls)) di1/dt ---> eq 4

Compare eq 1 and eq 4

Lp = (Lp' + (M*M /Ls)).
M^2 = 2.

K^2 = M^2 / (L1*L2),
K^2 = 2/ (4*2),
K^2 = 1/4,
K = 1/2 .
K= 0.5 (Answer).

Please explain how coefficient of coupling is 0.5.
When secondary open in both coil currents are going into coil.
i.e. Lp+M = 4.
And when secondary short circuited current in primary is going into coil and current in secondary coming out of coil,
i.e.
Lp-M =3.

From above Lp=3.5 and M= 0.5.
And coefficient of coupling is -
k= M/(Lp*Ls)^1/2.
k= 0.5/(3.5*2)^1/2.
k=0.188.