Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 74)
74.
Which transformer turns ratio is needed to match two 16-ohm parallel devices to an output of 22.2 ohms?
Discussion:
5 comments Page 1 of 1.
Wiblack. said:
5 years ago
Formula of relation between the primary and secondary:
((Np)^2)(Zp)=((Ns)^2)(Zs).
Since the primary (Zp) has two 16-ohm parallel it would result to 8-ohms in total (Zp = 8ohms).
Substituting through trial and error given by choices:
((1)^2)(8)=((0.6)^2)(22.2).
8 = 0.60 * 0.60 * 22.2.
8 = 7.992.
((Np)^2)(Zp)=((Ns)^2)(Zs).
Since the primary (Zp) has two 16-ohm parallel it would result to 8-ohms in total (Zp = 8ohms).
Substituting through trial and error given by choices:
((1)^2)(8)=((0.6)^2)(22.2).
8 = 0.60 * 0.60 * 22.2.
8 = 7.992.
Sandhyarani nayak said:
9 years ago
I can not understand. Please explain in detail.
KIRAN V said:
9 years ago
Zp = n^2 * Zs.
8 = 0.60 * 0.60 * 22.2.
8 = 7.992.
8 = 0.60 * 0.60 * 22.2.
8 = 7.992.
KIRAN V said:
9 years ago
Zp = n^2 * Zs.
Zp = Parallel combination of two 16 Ohm resistors.
Given,
Zs = Series Impedence/ Resistance = 22.2 Ohm.
n = no.of turns = 0.60.
Zp = Parallel combination of two 16 Ohm resistors.
Given,
Zs = Series Impedence/ Resistance = 22.2 Ohm.
n = no.of turns = 0.60.
Nisar khan said:
10 years ago
Zp = n^2 * Zs.
Here Zp is parallel combination of two 16 ohm R,
So their resultant is 8. Now put vales in equation you will get n = 0.6003.
So for matching 22.2*0.60 = 8 ohm.
Here Zp is parallel combination of two 16 ohm R,
So their resultant is 8. Now put vales in equation you will get n = 0.6003.
So for matching 22.2*0.60 = 8 ohm.
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