Solution to UniPhysics_II_HW #4

Solution to UniPhysics_II_HW #4 - Solution to -HW...

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Solution to -- HW assignment #4 Chapter 31: Electric Current and Resistance Note: you may need to use the following simplified relations to solve the following HW problems 1) dt dQ I = , 2) E j σ = , 3) Ed V = , 4) ρ 1 = 5) A I j / = , 6) A L R = --------------------------------------------------------------------------------------------------------------- Solve: (a) The current associated with the moving film is the rate at which the charge on the film moves past a certain point. The tangential speed of the film is () rev 1 min 2 rad 90 rpm 4.0 cm 90 1.0 cm 9.425 cm/s min 60 s 1 rev vr π ω == = × × × = In 1.0 s the film moves a distance of 9.425 cm. This means the area of the film that moves to the right in 1.0 s is (9.425 cm)(4.0 cm) = 37.7 cm 2 . The amount of charge that passes to the right in 1.0 s is Q = (37.7 cm 2 )( 2.0 × 10 9 C/cm 2 ) = 75.4 × 10 9 C Since IQt , we have ( ) 9 75.4 10 C 75.4 nA 1 s I −× The current is 75 nA. (b) Having found the current in part (a), we can once again use = Δ to obtain Δ t : 6 9 10 10 C 133 s 75.4 10 A Q t I Δ= = = × ---------------------------------------------------------------------------------------------------------------------------------- Solve: (a) The values of I (A) for selected values of t are t ( μ s) 0 1 2 4 6 8 10 I (A) 2.0 1.21 0.74 0.27 0.10 0.036 0.014 (b) Because Id Q d t = , 2.0 s 2.0 s 2.0 s 0 0 2.0 A 4.0 A s 4.0 C 1 t t t tt QI d t e d t e e μμ −− ⎤⎡ = = ⎦⎣ ∫∫ where we have used the condition Q = 0 C at t = 0 μ s. (c) The values of Q ( μ C) for selected values of t are t ( s) 0 1 2 4 6 8 10 Q ( C) 0 1.57 2.52 3.46 3.80 3.92 3.98
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---------------------------------------------------------------------------------------------------------------------------------- Solve: (a) The electric field inside the wire is wire E V L . Attaching the wire to the battery makes . Thus, wire V Δ= bat 1.5 V V 1.5 V 10 V/m 0.15 m E == (b) Using Table 31.2, the current density is 62 6 10 V/m 6.7 10 A/m 1.5 10 m E JE σ ρ = = × ×Ω (c) The current in a wire is related to the potential difference by wire IVR . Thus, wire 1.5 V 0.75 2 A V R I Δ = ==Ω The resistance is related to the wire’s geometry by 2 L L R Ar ρρ π () 6 4 1.5 10 m 0.15 m 3.1 10 m 0.31 mm 0.75 L r R ππ = × = Ω Thus, the wire’s diameter is d = 2 r = 0.62 mm. ------------------------------------------------------------------------------------------------------------------------------------------------------------ Solve: (a) The charge delivered is ( )( ) 36 50 10 A 50 10 s 2.5 C ×× = . (b) The current in the lightning rod and the potential drop across it are related by Equation 31.22. Using for iron from Table 31.2, ( ) ( ) 83 42 9.7 10 m 5.0 m 50 10 A 2.43 10 m 100 V AL I IV A LV × = = = × Δ This is the area required for a maximum voltage drop of 100 V. The corresponding diameter of the lightning rod is 3 8.8 10 m 8.8 mm A r × = × = - --------------------------------------------------------------------------------------------------------------------------------- Chapter 32: DC Circuit ----------------------------------------------------------------------------------------------------------------------------------
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Visualize: Solve: Despite the diagonal orientation of the 12 Ω resistor, the 6 Ω , 12 Ω , and 4 Ω resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is 1 111
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This note was uploaded on 04/21/2011 for the course ENC 3240 taught by Professor Meyers,p during the Spring '08 term at W. Florida.

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Solution to UniPhysics_II_HW #4 - Solution to -HW...

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