E37-1
The frequency, f , is related to the angular frequency ω by ω = 2πf = 2π(60 Hz) = 377 rad/s
The current is alternating because that is what the generator is designed to produce. It does this through the configuration of the magnets and coils of wire. One complete turn of the generator will (could?) produce one “cycle”; hence, the generator is turning 60 times per second. Not only does this set the frequency, it also sets the emf, since the emf is proportional to the speed at which the coils move through the magnetic field. E37-2 (a) XL = ωL, so f = XL /2πL = (1.28×103 Ω)/2π(0.0452 H) = 4.51×103 /s. (b) XC = 1/ωC, so C = 1/2πf XC = 1/2π(4.51×103 /s)(1.28×103 Ω) = 2.76×10−8 F. (c) The inductive reactance doubles while the capacitive reactance is cut in half. √ E37-3 (a) XL = XC implies ωL = 1/ωC or ω = 1/ LC, so p ω = 1/ (6.23×10−3 H)(11.4×10−6 F) = 3750 rad/s. (b) XL = ωL = (3750 rad/s)(6.23×10−3 H) = 23.4 Ω (c) See (a) above. E37-4 (a) im = E/XL = E/ωL, so im = (25.0 V)/(377 rad/s)(12.7 H) = 5.22×10−3 A. (b) The current and emf are 90◦ out of phase. When the current is a maximum, E = 0. (c) ωt = arcsin[E(t)/E m ], so ωt = arcsin
(−13.8 V) = 0.585 rad. (25.0 V)
and i = (5.22×10−3 A) cos(0.585) = 4.35×10−3 A. (d) Taking energy. E37-5 (a) The reactance of the capacitor is from Eq. 37-11, XC = 1/ωC. The AC generator from Exercise 4 has E = (25.0 V) sin(377 rad/s)t. So the reactance is XC =
1 1 = = 647 Ω. ωC (377 rad/s)(4.1µF)
The maximum value of the current is found from Eq. 37-13, im =
(∆VC )max ) (25.0 V) = = 3.86×10−2 A. XC (647 Ω)
(b) The generator emf is 90◦ out of phase with the current, so when the current is a maximum the emf is zero. 151