hw11 - Homework 11 Solutions Spring 2010 Required Problems...

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Homework 11 Solutions Spring 2010 Required Problems Chapter 26 44. ( a )From Eq. 26-7 the product RC is equal to the time constant. 6 9 3 24.0 10 s 1.60 10 F 15.0 10 RC C R τ τ - - = = = = y ( b )Since the battery has an EMF of 24.0 V, if the voltage across the resistor is 16.0 V, the voltage across the capacitor will be 8.0 V as it charges. Use the expression for the voltage across a charging capacitor. ( 29 ( 29 / / 6 6 1 1 ln 1 8.0V ln 1 24.0 10 s ln 1 9.73 10 s 24.0V t t C C C C V t V V e e V t τ τ τ τ - - - - = - = - - = - = - - = - - =0 e e e e 45. The current for a capacitor-charging circuit is given by Eq. 26-8, with R the equivalent series resistance and C the equivalent series capacitance. eq eq eq t R C I e R - = e ( 29 ( 29 eq 1 2 1 2 eq eq 1 2 1 2 ln ln IR I R R C C t R C R R C C + = - = - + + e e ( 29 ( 29 ( 29 ( 29 ( 29 2 6 3 3 6 3.8 10 F 1.50 10 A 4400 4400 ln 5.0 10 s 7.6 10 F 12.0V - - - - = - = � � Chapter 29 40. Rms voltage is found from the peak induced emf. Peak induced emf is calculated from Eq. 29-4. peak NB A ϖ = e + + + 1 R 1 C 2 C 2 R S e I
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( 29 ( 29 ( 29 ( 29 ( 29 2 peak rms 250 0.45T 2 rad rev 120rev s 0.050m 2 2 2 471.1V 470V NB A V π π ϖ = = = = e To double the output voltage, you must double the rotation frequency to 240 rev/s. 62. ( a ) From the efficiency of the transformer, we have S P 0.85 P P = . Use this to calculate the current in the primary. ( 29 S S P P P P P 75W 0.85 0.85 0.8021A 0.80A 0.85 0.85 110V P P P I V I V = = = = = ( b )The voltage in both the primary and secondary is proportional to the number of turns in the respective coil. The secondary voltage is calculated from the secondary power and resistance since 2 P V R = . ( 29 ( 29 P P P S S S S 110V 8.2 75W 2.4 N V V N V P R = = = = 79. ( a ) As the loop falls out of the magnetic field, the flux through the loop decreases with time creating an induced emf in the loop. The current in the loop is equal to the emf divided by the resistance, which can be written in terms of the resistivity using Eq. 25-3. 2 2 2 / 4 4 16 16 B d d d dA d I B B v R dt dt π π π ρ ρ ρ Φ = = = = e l l l l This current induces a force on the three sides of the loop in the magnetic field. The forces on the two vertical sides are equal and opposite and therefore cancel. 2 2 2 16 16 d d B v F I B B v B π π ρ ρ = = = l l l l l By Lenz’s law this force is upward to slow the decrease in flux. ( b )Terminal speed will occur when the gravitational force is equal to the magnetic force. 2 2 2 2 16 4 4 16 m T g m T g d B v d F g v B ρρ π ρ π ρ = = = l l ( c ) We calculate the terminal velocity using the given magnetic field, the density of copper from Table 13-1, and the resistivity of copper from Table 25-1 ( 29 ( 29 ( 29 ( 29 3 3 8 2 2 16 8.9 10 kg/m 1.68 10 m 9.80 m/s 3.7 cm/s 0.80 T T v - = = Chapter 30
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2. If we assume the outer solenoid is carrying current 1 , I then the magnetic field inside the outer solenoid is 0 1 1 . B n I μ = The flux in each turn of the inner solenoid is 2 2 21 2 0 1 1 2 . B r n I r π μ π Φ = = The mutual inductance is given by Eq. 30-1.
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