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# solutions_36 - PHY2061 R D Field Chapter 36 Solutions C R I...

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PHY2061 R. D. Field Solutions Chapter 36 Page 1 of 9 Chapter 36 Solutions Problem 1: A current I flows through a resistor with R=25 k , a capacitor with C=4 milliF , and an inductor with L=7 Henry as shown in the Figure. At a time when Q=8 milliC and the current I=0.2 milliAmps and is changing at a rate dI/dt=1.0 A/s , what is the potential difference V B -V A ( inVolts )? Answer: -14 Solution: Moving in the direction of the current flow we have VV I R Q C L dI dt BA −= , and V s A H dt dI L V mF mC C Q V A IR 7 ) / 1 )( 7 ( 2 4 8 5 ) 10 25 )( 10 2 . 0 ( 3 3 = = = = = × × = Thus, V B -V A = -5V -2V -7V = -14V. R I A B L 1 L 2 Problem 2: A current I flows through a resistor with R = 333 , and two inductors in parallel with L 1 = 2 Henry and L 2 = 4 Henry as shown in the Figure . At a time when the current I = 1.0 milliAmps and is changing at a rate dI/dt = -1.0 A/s , what is the potential difference V B -V A ( in Volts )? Answer: 1 C R I AB L + - Q

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PHY2061 R. D. Field Solutions Chapter 36 Page 2 of 9 Solution: Moving in the direction of the current flow we have dt dI L IR V V eff A B = , where H H H L L L eff 4 3 4 1 2 1 1 1 1 2 1 = + = + = , since for inductors in parallel you add the inverses. Thus, L eff = 1.33 H and V s A H dt dI L V A IR eff 33 . 1 ) / 1 )( 33 . 1 ( 333 . 0 ) 333 )( 10 1 ( 3 = = = × = so that , V B -V A = -0.333V +1.33V = 1V. Problem 3: What is the magnitude of the uniform magnetic field ( in Tesla ) that contains as much energy per unit volume as a uniform 3,000 V/m electric field? Answer: 1x10 -5 Solution: Setting the magnetic field energy density equal to the electric field energy density yields 2 0 2 0 2 1 2 1 E u B u E B ε µ = = = and solving for B gives T s m m V c E E B 5 8 0 0 10 1 / 10 3 / 000 , 3 × = × = = = .
PHY2061 R. D. Field Solutions Chapter 36 Page 3 of 9 C R I AB L 1 + - Q L 2 Problem 4: A current I flows through a resistor with R = 2 , a capacitor with C = 3 milliF , and and two inductors in parallel ( L 1 = 1 milliHenry and L 2 = 3 milliHenry ) as shown in the Figure . At a time when Q = 0.6 microC and the current I = 0.2 milliAmps and is decreasing at a rate of 0.8 Amps/s , what is the potential difference V B -V A ( milliVolts )?

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solutions_36 - PHY2061 R D Field Chapter 36 Solutions C R I...

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