quiz_solutions09

# Fundamentals of Physics, (Chapters 21- 44) (Volume 2)

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1213 PY212 Solutions Quiz 9 (a) The circuit consists of one generator across one inductor; therefore, V max = V L . The current amplitude is A. 955 . 0 H) 40 rad/s)(0.0 14 3 ( V 0 . 12 max max = = = = L V X V I d L ω (b) When the current is at a maximum, its derivative is zero. Thus, Eq. 30-35 gives V L = 0 at that instant. Stated another way, since V ( t ) and i ( t ) have a 90° phase difference, then V ( t ) must be zero when i ( t ) = I . The fact that φ = 90° = π /2 rad is used in part (c). (c) To find the current i ( t ), we must first determine at what time t the emf is 6.0 V and increasing in magnitude. Consider V ( t ) = V max sin( t ), with V ( t ) = 6.0 V. In order to satisfy this equation, we require that sin( t ) = 1/2. This actually gives two solutions:
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Unformatted text preview: t = π / 6 and t = 5 / 6. The problem states that the emf is increasing, so the slope of V must be positive at time t . The slope is given by the derivative of V ( t ): ) cos( ) ( max t V t V dt d = For t = / 6 we obtain cos( t ) > 0, and for t = 5 / 6 we get cos( t ) < 0. Evidently, the right solution is t = / 6. To find the current at this time, we simple substitute it into Eq. 31-52: A. 827 . ) 2 / 6 / sin( ) A 955 . ( ) sin( ) ( − = − = − = π t I t i L Hence, the current is 0.827 A at that moment....
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