The little solenoid electromagnet has 1200 loops of thin wire - the big solenoid
ID: 1262286 • Letter: T
Question
The little solenoid electromagnet has 1200 loops of thin wire - the big solenoid (connected to the galvanometer) has 560 loops, each 70mm diameter. Because the iron rod is longer than the 150mm big solenoid, about 80% of its magnetic flux returns to its South end outside the big solenoid coils. Suppose I plug the little solenoid directly into a 110VRMS at 60 cycle/second wall outlet ...(show steps!) . . . a) compute the peak Voltage at the plug's positive terminal . . . and the angular frequency of the voltage wave-form (in radian/s). . . . b) compute the little solenoid's Inductive reactance at this angular frequency . . . recall that R = 30? and L = 0.090 Henry. . . . c) add its Inductive Impedance to its Resistive Impedance , to find its total Impedance . . . then find the maximum current when plugged in. . . . d) calculate the & from the big solenoid (specify RMS or max) ... why can't we calculate its current?Explanation / Answer
Given that -
Vrms=110V
(A) we know that peak voltage Vp= ?2*Vrms=1.414*110=155.54V.
And for angular frequency it is given that w=60cycle/sec we know that 1cycle=2pi
hence w=60*2*3.14=376.8 rad/sec.
(B) The little solenoid's Inductive reactance XL= wL=33.912ohm.
(C) Total Impedance = 30ohm+33.912ohm=63.912ohm.
total current i=V/total impedance=155.54/63.912=2.4A.
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