Answer to Question #277098 in Electricity and Magnetism for jojo

Question

transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 Ω

Ω

.

(a) What should the ratio of primary to secondary turns of the transformer be?

(b) What rms current must the secondary supply?

(c) What average power is delivered to the load?

(d) What resistance connected directly across the 120-V line would draw the same power as the transformer? Show that this is equal to 5.00 Ω

Ω

 times the square of the ratio of primary to secondary turns.

Answer 1

a)

$k=\frac{U_1}{U_2}=\frac{N_1}{N_2}=10,$

b)

$I_2=\frac{U_2}{R_2}=0.24~A,$

c)

$P=P_1=P_2=U_2I_2=2.88~W,$

d)

$P=\frac{U_1^2}{R_1}=U_2I_2,\implies$

$R_1=\frac{U_1^2}{U_2R_2}=\frac{U_1^2U_2}{U_2^2I_2}=R_2k^2=5~k\Omega.$