FIGURE 4(a) shows two inductances connected in parallel across an a.c. supply.
(a) Apply Kirchhoff’s voltage law to loop abef and to loop abcdef of the circuit.
(b) Hence or otherwise obtain the current ratio I1/I2 in terms of the circuit inductances.
(c) See if you can show that L1 and L2 can be replaced by the
equivalent inductor, Leq, of FIGURE 4(b) where Leq = (L1*L2-M^2)/(L1+L2-2*M)
(d) A 1 nF capacitor is placed across the two inductors (FIGURE 4 (c)). If L1 = L2 = L and k = 0.5, determine the required value of L if the minimum current I flows from the supply when it is at a frequency of 1 MHz.
The Attempt at a Solution
(a) ABEF : V=jw*L1*I1+jw*M*I2, ABCDEF : V=jw*L2*I2+jw*M*I1
(b) I1= (V*(L2-M)) / (jw*(L1*L2-M^2)) , I2=(V*(L1-M)) / (jw*(L1*L2-M^2))
I1/I2= (((V*(L2-M)) / (jw*(L1*L2-M^2))) / (((V*(L1-M)) / (jw*(L1*L2-M^2)))
I1/I2 =L2-M / L1-M
(c) Leq = I1+I2 , = (((V*(L2-M)) / (jw*(L1*L2-M^2))) + (((V*(L1-M)) / (jw*(L1*L2-M^2)))
=(V*(L1+L2-2M))/(jw*(L1*L2-M^2)) = V/jw*Leq, Leq = (L1*L2-M^2)/(L1+L2-2*M)
(d) Xc= 1/(2*pi*f*C) = j159.155 ohms
The bit i'm stuck on is whether I can solve for L using :
However this is for resonant frequency. A push in the right direction anyone on any of these questions. Can someone explain the origin of why in (c) I1+I2= V/jw*Leq
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