{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework3 - part is connected across an AC voltage source...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 334, Winter Quarter 2012 Electric Circuits Laboratory I Homework Assignment 3 Due Tuesday January 31 Note: Sloppy work will not be graded and will receive zero points. 1. A R-C high-pass filter contains a 1 μ F capacitor and a 1k resistor. Suppose a single “square” voltage pulse of 5 V for 1 ms duration is applied to the input. Carefully and neatly sketch the resulting output, including numeric values, of output voltage versus time. 2. A two-terminal “black box” is known to contain an inductor L, a capacitor C, and a resistor R. On connecting a 1.5 V battery, 1.5 mA flows. When an AC voltage of 1V RMS at 60 Hz is connected, 10 mA RMS flows. As the frequency increases at a fixed 1 V RMS, the current reaches a maximum of over 100 A at 1 kHz. Carefully and neatly sketch the circuit in the black box and find values for L, C and R. 3. A load impedance 1000(1 + i) Ω (and note it contains an imaginary
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: part) is connected across an AC voltage source of amplitude 10 V and frequency 60 Hz. What ʼ s the power dissipated over one cycle within the load? Comment: There are several ways to attack this problem. For definiteness, my solution started by choosing a specific phase of the voltage so it ʼ s a maximum at t=0. Then I wrote everything in complex form. For those not familiar with complex analysis: To find power, I used the identity that the time-average is ½ the real part of the product of a complex current amplitude times the complex conjugate of the voltage amplitude. See the discussion in the textbook pp. 33-34 (especially the lower left column on p 34: notice the text speaks of “complex RMS amplitudes”) while my factor of ½ comes from using the full complex amplitudes. You can also of course find the time average by directly averaging over a complete cycle....
View Full Document

{[ snackBarMessage ]}