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# 29 - Kapoor(mk9499 oldhomework 29 Turner(60230 This...

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Kapoor (mk9499) – oldhomework 29 – Turner – (60230) 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. None of these 5. R E C L 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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Kapoor (mk9499) – oldhomework 29 – Turner – (60230) 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
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