CHAPTER 22 ELECTROMAGNETIC INDUCTION

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Melvin Summers
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1 CHAPTER 22 ELECTROMAGNETIC INDUCTION PROBLEMS 47. REASONING AND Using Equation 22.7, we find emf 2 M I or M ( emf 2 ) t ( 0.2 V) ( 0.4 s) t I (.6 A) ( 3.4 A) H 49. SSM REASONING AND From the results of Example 3, the self-inductance L of a long solenoid is given by L µ 0 n 2 Al. Solving for the number of turns n per unit length gives n L µ 0 Al H (4π 0 7 T m/a)( m 2 )(0.052 m) turns/m Therefore, the total number of turns N is the product of n and the length l of the solenoid N nl ( turns/m)(0.052 m) 220 turns 54. REASONING AND The mutual inductance is The flux through the 25 turn coil is M N 2 Φ 2 /I so Φ 2 B A 2 µ o n I A 2 M N 2 µ o n A 2 (25)(4p 0 7 T. m/a)(750/m)(π)(0.080 m) H 6. SSM REASONING AND 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 is given by the transformer equation, Equation 22.2: N s N p V s V p 0.0 V 20 V 2 so the turns ratio is :2.

2 684 ELECTROMAGNETIC INDUCTION 65. SSM REASONING The power lost in heating the wires is given by Equation 20.6b: P I 2 R. Before we can use this equation, however, we must determine the total resistance R of the wire and the current that flows through the wire. a. The total resistance of one of the wires is equal to the resistance per unit length multiplied by the length of the wire. Thus, we have ( Ω/km)(7.0 km) 0.35 Ω and the total resistance of the transmission line is twice this value or 0.70 Ω. According to Equation 20.6a ( P IV ), the current flowing into the town is Thus, the power lost in heating the wire is I P V.2 06 W 200 V.0 03 A P I 2 R ( A) 2 ( 0.70 Ω) W b. According to the transformer equation (Equation 22.2), the stepped-up voltage is V s V p N s N p ( 200 V) V According to Equation 20.6a (P IV), the current in the wires is The power lost in heating the wires is now I P V W V.0 0 A P I 2 R (.0 0 A) 2 ( 0.70 Ω) W

3 CHAPTER 23 ALTERNATING CURRENT CIRCUITS Chapter 22 Problems 685 PROBLEMS 3. REASONING AND a. The equivalent capacitance of capacitors wired in parallel is b. The capacitive reactance is C p C + C 2 + C µf µf µf 3.0 µf X C Therefore, the current is 2π fc p 2π (440 Hz)( F) 28 Ω I V X C 7V 28 Ω 0.6 A 6. REASONING The capacitance C is related to the capacitive reactance X C and the frequency f via Equation 23.2 as C /(2π f X C ). The capacitive reactance, in turn is related to the rms-voltage V rms and the rms-current I rms by X C V rms /I rms (see Equation 23.). Thus, the capacitance can be written as C I rms /(2π f V rms ). The magnitude of the maximum charge q on one plate of the capacitor is, from Equation 9.8, the product of the capacitance C and the peak voltage V. a. Recall that the rms-voltage V rms is related to the peak voltage V by V rms is, then, V 2. The capacitance C I rms 2π fv rms 3.0A 2π 750 Hz ( ) 40 V F b. The maximum charge on one plate of the capacitor is q CV ( F)40 ( V) C 0. REASONING AND We know that V IX L I(2π f L) (0.20 A)(2π)(750 Hz)(0.080 H) 75 V

4 686 ELECTROMAGNETIC INDUCTION 4. REASONING AND First, we calculate the inductive reactances: X L 2π f L 2π(235 Hz)(0.460 H) 679 Ω X L2 2π f L 2 2π(235 Hz)(0.85 H) 273 Ω When wired in parallel, both inductors have the same voltage across them, V IX L, so that I V/X L (2.0 V)/(679 Ω) A I 2 V/X L2 (2.0 V)/(273 Ω) A The total current the generator delivers is, therefore, I I + I A A A 7. SSM REASONING The voltage supplied by the generator can be found from Equation 23.6, V rms I rms Z. The value of I rms is given in the problem statement, so we must obtain the impedance of the circuit. The impedance of the circuit is, according to Equation 23.7, Z R 2 +( X L X C ) 2 (275 Ω) 2 +(648 Ω 45 Ω) Ω The rms voltage of the generator is V rms I rms Z (0.233 A)( Ω) 83.9 V 24. REASONING AND a. At very large frequencies the impedance of the inductor becomes much larger than that of the capacitor or the resistances of the two resistors. Therefore, current flows only through the top branch of the circuit, i.e., through the capacitor and 290 W resistor. For a very large frequency, X C 0, so Z R 290 W. Therefore, I V/Z (75 V)/(290 W) 0.26 A b. At very small frequencies the impedance of the capacitor becomes much larger than that of the inductor or the resistances of the two resistors. Therefore, current flows only through the middle branch of the circuit, i.e., through the inductor and 70 W resistor. At very small frequencies, X L 0, so Z R 70 W. Therefore, I V/Z (75 V)/(70 W) 0.A

5 Chapter 22 Problems REASONING The instantaneous value of the generator voltage is given by V (t) V 0 sin 2π ft, where V 0 is the peak voltage and f is the frequency. We will see that the inductive reactance is greater than the capacitive reactance, X L > X C, so that the current in the circuit lags the voltage by π/2 radians, or 90. Thus, the current in the circuit obeys the relation I(t) I 0 sin (2π ft π/2), where I 0 is the peak current. a. The instantaneous value of the voltage at a time of s is V (t ) V 0 sin 2π ft [ ( )] 5.8 V ( 64.0 V)sin 2π( Hz) s Note: When evaluating the sine function in the expression above, be sure to set your calculator to the radian mode. b. The inductive and capacitive reactances are X L X C 2π fl 2π( Hz)4.30 ( 0 3 H) Ω 2π fc 2π( Hz) F ( ) 8.08 Ω Since X L is greater than X C, the current lags the voltage by π/2 radians. Thus, the instantaneous current in the circuit is I(t) I 0 sin (2π ft π/2), where I 0 V 0 /Z. The impedance Z of the circuit is Z R 2 + ( X L X C ) ( 27.0 Ω 8.Ω) Ω The instantaneous current is I V 0 Z sin ( 2π ft 2 π ) 64.0 V [ ] 4.2 A 8.94 Ω sin 2π ( Hz).50 ( 0 4 s) 2 π 29. SSM REASONING The current in an RCL circuit is given by Equation 23.6, I rms V rms / Z, where the impedance Z of the circuit is given by Equation 23.7 as Z R 2 + (X L X C ) 2. The current is a maximum when the impedance is a minimum for a given generator voltage. The minimum impedance occurs when the frequency is f 0, corresponding to the condition that X L X C, or 2π f 0 L /( 2π f 0 C). Solving for the frequency f 0, called the resonant frequency, we find that f 0 2π LC Note that the resonant frequency depends on the inductance and the capacitance, but does not depend on the resistance.

6 688 ELECTROMAGNETIC INDUCTION a. The frequency at which the current is a maximum is f 0 2π LC 352 Hz 2π ( H)( F) b. The maximum value of the current occurs when f f 0. This occurs when X L X C, so that Z R. Therefore, according to Equation 23.6, we have I rms V rms Z V rms R 55 V 0.0 Ω 5.5 A 37. SSM WWW REASONING Since we know the values of the resonant frequency of the circuit, the capacitance, and the generator voltage, we can find the value of the inductance from Equation 23.0, the expression for the resonant frequency. The resistance can be found from energy considerations at resonance; the power factor is given by cosφ, where the phase angle φ is given by Equation 23.8, tanφ (X L X C )/ R. a. Solving Equation 23.0 for the inductance L, we find that L 4π 2 f 2 0 C 4π 2 ( Hz) 2 ( F) H b. At resonance, f f 0, and the current is a maximum. This occurs when X L X C, so that Z R. Thus, the average power P provided by the generator is P V 2 rms / R, and solving for R we find R V 2 rms (.0 V)2 P 25.0 W 4.84 Ω c. When the generator frequency is 2.3 khz, the individual reactances are X C The phase angle φ is, from Equation 23.8, 2π fc 2π( Hz)( F) 3.5 Ω X L 2π fl 2π( Hz)( H) 42.7 Ω φ tan The power factor is then given by X L X C R tan 42.7 Ω 3.5 Ω 4.84 Ω cosφ cos

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