Homework_Ch_24_solutions

# Homework_Ch_24_solutions - P202 Spring 2006 Ch 24 ac...

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Unformatted text preview: P202 Spring 2006 Ch 24: ac Circuits Homework due at 11:59 pm Feb. 26, 2006 Error allowance on all numerical answers is 3%. 1. The relationship 2 / max V V rms = is valid only for voltages that vary sinusoidally. What is the rms voltage for the square wave voltage shown below?__________ V Solution: The definition of root-mean-square is “square ROOT of the MEAN value of the SQUARE of a quantity”. The square of the voltage sketched above is 25 V 2 for all values of t (time). The square root of 25 is 5. So the rms voltage is 5 V. Solution: 5 2. An ac generator with a frequency of F=[20,25,30 or 35] Hz and an rms voltage of 15 V is connected to a C=[integer between 30 & 40] µ F capacitor. Assume that the generator produces a sinusoidal waveform. (a) What is the maximum current in the circuit? ________________mA (b) What is the current in the circuit when the voltage across the capacitor is 7.5 V and increasing? ____________ (c) What is the current in the circuit when the voltage across the capacitor is 7.5 V and decreasing? _____________ Solution: We know how to find the rms current in the circuit given the rms voltage and the capacitive reactance. We use the generalized form of Ohm’s law given in eqn 24-8 of Walker. The maximum current in the circuit is related to the rms current by the usual sqrt(2) factor for a sinusoidal voltage (eqn 24-5 of Walker). To solve parts (b) and (c) it helps to draw phasor diagrams for a circuit with a capacitor (above) for various times, t. Note that ω t is the angle between the x axis and the current in all cases even though the angle is only marked on the leftmost figure for clarity. Remembering that the value of a quantity represented by a phasor is just the component of the phasor along the y axis, we see from the leftmost diagram that the current is given by t I I ω sin max = and the voltage across the capacitor is given by ) 2 / sin( max π ω − = t V V . This is just another way of saying that the voltage across a capacitor lags behind the current. From the second phasor diagram we can see that the voltage is positive (i.e. the y component of the voltage phasor is positive) and increasing when ω t is between π /2 and π and that in this case the current is positive and decreasing (remember that both phasors rotate counter clockwise). From the third figure, we see that the voltage is positive and decreasing when ω t is between π and 3 π /2 and that in this case the current is negative....
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Homework_Ch_24_solutions - P202 Spring 2006 Ch 24 ac...

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