# Average power to ZL coupling coefficent and mutual inductance M?

1. Jun 12, 2013

### asdf12312

1. The problem statement, all variables and given/known data

In the circuit of Figure 4-2, let N1=2000 turns, N2=1000 turns, R=0, Z1=0 and ZL=300+j400 (Ω). If V1(max)=1500 V, what is the average power delivered to ZL?
A. 5400 W
B. 2700 W
C. 337.5 W
D. 675 W

2. Relevant equations
V2=(N2/N1)V1

3. The attempt at a solution
since turn factor is 2:1 then V2= (1/2)V1=750V. and I2=750/(300+j400)= 0.9-1.2i=1.5A (magnitude of polar form)

here is where i get confused. my book for some examples uses (1/2)I2R for power, however i also know the general power equation is I2R. in this case i decided to go with the book's example so to find average power to ZL I used (1/2)I2R and only used resistive part of ZL in calculating power:
1/2 (1.5)2*(300) = 337.5. so i put C as my answer.

Last edited: Jun 12, 2013
2. Jun 12, 2013

### Staff: Mentor

The 1/2 comes from taking the RMS value of the given peak value. So $I_{RMS} = I_{peak}/\sqrt{2}$, and squaring it yields a factor of (1/√2)2 = 1/2.