Practice Problem - Chapter 22, Part 2

35. SSM REASONING We can use the information given in the problem statement to determine the area of the coil A. Since it is square, the length of one side is g1g1 g65 = A . SOLUTION According to Equation 22.4, the maximum emf 0 ? induced in the coil is 0 NAB? ?= . Therefore, the length of one side of the coil is 0 75.0 V 0.150 m (248)(0.170 T)(79.1 rad/s) A NB ? ? == = =g65 ________________________________________________________________________ ______ 44. REASONING When the current through an inductor changes, the induced emf ? is given by Equation 22.9 as = I L t ? ? ? ? where L is the inductance, ?I is the change in the current, and ?t is the time interval during which the current changes. For each interval, we can determine ?I and ?t from the graph. SOLUTION a. () 3 3 4.0 A 0 A = 3.2 10 H 6.4 V 2.0 10 s 0 s I L t ? ? ? g167g183?? ?=?× =? g168g184 ? ×? g169g185 b. () 3 33 4.0 A 4.0 A =3.210H 0V 5.0 10 s 2.0 10 s I L t ? ? ?? g167g183 ?=?× = g168g184 ? ×?× g169g185 c. () 3 33 0A 4.0A = 3.2 10 H 3.2 V 9.0 10 s 5.0 10 s I L t ? ? ?? g167g183?? ?=?× =+ g168g184 ? ×?× g169g185 ________________________________________________________________________ ______ 49. SSM REASONING The rate of change g39I P /g39t in the primary current is related to the emf S ? induced in the secondary coil according to SP /MI t? =? ? ? (Equation 22.7), where M is the mutual inductance between the coils. We can use this expression to determine g39I P directly. However, in doing so, we will omit the minus sign, since the direction of the current is unspecified. A value for the induced emf can be obtained from the induced current I S and the resistance R in the secondary circuit, according to SS I R? = , which is Ohm?s law. SOLUTION Substituting the induced emf from Ohm?s law into Equation 22.7 (without the minus sign), and solving for g39I P gives ()() ()() 33 SP SS P 3 6.0 10 A 12 72 10 s or 1.6 A 3.2 10 H IR tI IR M I tM ? ?? ? ×?× ?? == ?= = = ? × 56. REASONING The ratio ()sp /I I of the current in the secondary coil to that in the primary coil is equal to the ratio ()p s /NN of the number of turns in the primary coil to that in the secondary coil. This relation can be used directly to find the current in the primary coil. SOLUTION Solving the relation ()( )sp p s //I INN= (Equation 22.13) for I p gives () s ps p 1 1.6 A 0.20 A 8 N II N g167g183 g167g183 == = g168g184g168g184 g168g184 g169g185 g169g185 ________________________________________________________________________ ______ 57. SSM REASONING AND SOLUTION Since the secondary voltage is less than the primary voltage, we can conclude that the transformer used in the doorbell described in the problem is a step - down transformer . The turns ratio N s /N p is given by the transformer equation, Equation 22.12: N s N p = V s V p = 10.0 V 120 V = 1 12 so the turns ratio is 1: 12 . ________________________________________________________________________ ______ 60. REASONING The power that your house is using can be determined from rms rms PI?= (Equation 20.15a), where I rms is the current in your house. We know that rms ? is 240 V. To find I rms , we must apply S P PS I N I N = (Equation 22.13) to convert the current given for the primary of the substation transformer into the current in the secondary of the substation transformer. This secondary current then becomes the current in the primary of the transformer on the telephone pole. We will use Equation 22.13 again to find the current in the secondary of the transformer on the telephone pole. This secondary current is I rms . SOLUTION According to Equation 20.15a, the power that your house is using is rms rms PI?= (1) Applying Equation 22.13 to the substation transformer, we find () 3S PP SP PS S Substation 29 or 48 10 A 1.4 A 1 I NN II IN N ? g167g183 g167g183 ===×= g168g184 g168g184 g168g184 g169g185 g169g185 g8g11g9g11g10 Thus, the current in the primary of the transformer on the telephone pole is 1.4 A. Applying Equation 22.3 to the telephone-pole transformer, we obtain () S PP SP PS S Telephone pole 32 or 1.4 A 45 A 1 I NN II IN N g167g183 g167g183 ==== g168g184 g168g184 g168g184 g169g185 g169g185 g8g11g9g11g10 Using this value of 45 A for the current in Equation (1) gives ()( ) 4 rms rms 45 A 240 V 1.1 10 WPI?== =× sw Microsoft Word - prob_035.doc