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# HW 31 - terry(ect328 homework 31 Turner(59130 This...

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terry (ect328) – homework 31 – Turner – (59130) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The current in a light bulb is plotted as a function of angular velocity ω as shown below. ω I light bulb Select the circuit which gives rise to the above plot. 1. R E C L 2. R E C L cor- rect 3. R E C L 4. R E C L 5. None of these Explanation: X C = 1 ω C and X L = ω L . Analyze each circuit: R E C L Z RC = radicalBig R 2 + X 2 C = radicalBigg R 2 + 1 ( ω C ) 2 I R = E radicalbigg R 2 + 1 ( ω C ) 2 ω I light bulb ω 0 ω 0 = 1 R C ====================== R E C L Z RL = radicalBig R 2 + X 2 L = radicalBig R 2 + ( ω L ) 2 I R = E radicalbig R 2 + ( ω L ) 2 ω I light bulb ω 0 ω 0 = R L ====================== R E C L

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terry (ect328) – homework 31 – Turner – (59130) 2 Z = R + 1 radicalBigg parenleftbigg 1 X C - 1 X L parenrightbigg 2 = R + 1 radicalBigg parenleftbigg ω C - 1 ω L parenrightbigg 2 I R = E R + 1 radicalBigg parenleftbigg ω C - 1 ω L parenrightbigg 2 ω I light bulb ω 0 ω 0 = 1 L C ====================== R E C L Z = radicalBig R 2 + ( X L - X C ) 2 = radicalBigg R 2 + parenleftbigg ω L - 1 ω C parenrightbigg 2 I R = E radicalBigg R 2 + parenleftbigg ω L - 1 ω C parenrightbigg 2 ω I light bulb ω 0 ω 0 = 1 L C 002 10.0 points A circuit containing a resistor and an in- ductor is shown below. R L b a Select the plot commensurate with the above circuit in which the impedance from a to b is plotted as a function of angular ve- locity ω .
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