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Transcript
Talk about the application of capacitors in flashbulbs or windshield wipers. How does a capacitor act when switch first closes until it reaches steady state or fully charged? When the switch first closes + - How does the current change over time? How does the voltage across the resistor change over time? How does the voltage across the capacitor change over time? + - During Transient Response + - Short At steady state + - Resistor Open How does an inductor act when switch first closes until it reaches steady state or fully charged? When the switch first closes How does the current change over time? How does the voltage across the resistor change over time? How does the voltage across the inductor change over time? + - During transient response + - + - Open At steady state + - Resistor Short Time constant (Tau) • The time required to charge a capacitor to 63.2% of maximum voltage, or the time to discharge a capacitor to 36.8% of its final voltage. = R∙C Seconds • Similarly for an inductor, it is the time required to change the amount of current through an inductor to 63.2% of max current (or reduce it to 36.8% of final amount) = L/R Seconds 95% 86.5% 98.2% 99.3% Every time constant, the voltage rises 63% of what is remaining. 63.2% See table 21.1 in your book for these values. Determining the time constant • What is the time constant of a 0.01uF capacitor in series with a 2kΩ resistor? • 20us • What is the time constant of a 10uF capacitor in series with a 100kΩ resistor? • 1 sec • What is the time constant for a 200mH inductor with a 2Ω resistor? • 100ms • What value of resistance is needed to cause a of 1.2ms with a 4.7uF capacitor? • 255 Ω Not quite so Random VOTD • http://www.youtube.com/watch?v=coW1RHUsf_I&feature=yo utube_gdata_player • Only watch til 2:45min • Shorting a capacitor: http://www.youtube.com/watch?v=gj1pkyCL75E 95% 86.5% 63.2% 98.2% 99.3% A 50kΩ resistor is connected in series with a 40uF capacitor. With a DC source of 50V, what is the charge across the capacitor after 6 sec? Assume VC=0 at t=0 = RC = 50k x 40u = 2sec 95% 86.5% 63.2% 98.2% 99.3% A 10kΩ resistor is connected in series with a .01uF capacitor. With a DC source of 20V, what is the charge across the capacitor after 200us? Assume VC=0 at t=0 = RC = 10k x .01u = 100us Do the following 2 problems on your own. • What is the voltage across a 5uF capacitor connected in series with a 22kΩ resistor after 330ms with a 30V DC source voltage? (Assume 0V for start up) • For the problem above, what is the voltage across the resistor after 440ms? Try one more… • What is the voltage across a 20uF capacitor connected in series with a 100kΩ resistor after 3s if the source voltage is 10V? (Assume 0V for start up) • Hint: It is not 74.85% or 7.485V • When the amount of time does not fall exactly on an even number of time constants, such as 1, 2, etc. then we use the following equation 21.2: 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 • • • • 𝑡 − 𝜏 VC is the voltage across the capacitor VS is the DC source voltage t is the amount of time elapsed is the time constant ) So what the heck is e in the𝑡 equation: 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 − 𝜏 ) • Everyone think of a large number. Something larger than 1000. • Now plug that number into the formula, where N is your number: • (1+1/N)N • With the help of magic I bet your number is: • 2.718… • Find the “e” on your calculator and press enter. • (There are 2 buttons, a green one and ex) Going back to our original problem • What is the voltage across a 20uF capacitor connected in series with a 100kΩ resistor after 3s if the source voltage is 10V? (Assume 0V for start up) • Hint: It is not 74.85% 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 • • • • VC = ? VS = 10V t=3 = 2sec 𝑉𝐶 = 10 (1 − 𝑒 𝑉𝐶 = 7.76𝑉 𝑡 − 𝜏 3 − 2 ) ) • What is the voltage across a .002uF capacitor connected in series with a 22kΩ resistor after 160us if the source voltage is 12V? (Assume 0V for start up) • • • • VC = ? VS = 12V t = 160us = 44us 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 𝑉𝐶 = 12 (1 − 𝑒 𝑡 − 𝜏 ) 160𝑢 − 44𝑢 ) 𝑉𝐶 = 11.68𝑉 Does this answer make sense? How many time constants have passed? Calculators are about to become very important in this class. • What is the voltage across a .05uF capacitor connected in series with a 500Ω resistor after 75us if the source voltage is 100V? (Assume 0V for start up) • • • • VC = ? VS = 100V t = 75us = 25us 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 𝑡 − 𝜏 𝑉𝐶 = 100 (1 − 𝑒 ) 75𝑢 − 25𝑢 ) 𝑉𝐶 = 95𝑉 Does this answer make sense? How many time constants have passed? RVOTD • http://www.youtube.com/watch?v=syCpfNu1Hqc • (no volume) Get to here before lab 30. • What is the voltage across a .05uF capacitor connected in series with a 500Ω resistor after 2.2 if the source voltage is 100V? (Assume 0V for start up) Doing it 2 different ways: = 25us 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 ) t = 2.2(25us) 𝑉𝐶 = 100 (1 − 𝑒 2.2(25𝑢) 25𝑢 ) 𝑉𝐶 = 100 (1 − 𝑒 − 𝑉𝐶 = 100 (1 − 𝑒 𝑉𝐶 = 100 (1 − 𝑒 −2.2 ) 𝑉𝐶 = 88.9𝑉 𝑡 − 𝜏 𝑉𝐶 = 88.9𝑉 − 2.2 ) −2.2 ) Solving for t: 𝑉𝐶 = 𝑉𝑆 (1 − 𝑒 − 𝑡 𝜏 ) • How much time does it take to charge a 4uF capacitor to 5V if there is a 10V DC source and a 10kΩ resistor in series? 𝑉𝐶 𝑡 = −𝜏 ∙ ln(1 − ) 𝑉𝑆 5 𝑡 = −40𝑚𝑠 ∙ ln(1 − ) 10 𝑡 = 27.7𝑚𝑠 Shall we have another… • How many time constants does it take to charge a capacitor to 25% of being fully charged? 𝑉𝐶 𝑡 = −𝜏 ∙ ln(1 − ) 𝑉𝑆 𝑡 = −𝜏 ∙ ln(1 − .25) 𝑡 = 0.288𝜏 Calculating Current in RC circuits • Since we know 𝑉𝐶 = 𝑉𝑆 (1 • To calculate VR it would be 𝑉 𝑡 − 𝜏 −𝑒 ) 𝑅 = 𝑉𝑆 − 𝑉𝐶 = 𝑉𝑆 − 𝑉𝑆 (1 − 𝑒 = 𝑉𝑆 (1 − 1 + 𝑒 = 𝑉𝑆 (𝑒 𝑡 −𝜏 • Using Ohms Law IR would be 𝑉𝑆 (𝑒 𝐼𝑅 = 𝑅 − ) 𝑡 𝜏 ) = 𝐼𝐶 𝑡 𝜏 𝑡 − 𝜏 − ) ) Thus… • 𝐼𝐶 = 𝑡 −𝜏 𝑉𝑆 (𝑒 𝑅 ) = 𝐼𝑅 𝑉𝑅 = 𝑅 What is the current through a .002uF capacitor connected in series with a 22kΩ resistor after 160us if the source voltage is 12V? (Assume 0V for start up) − 𝑉𝑆 (𝑒 𝐼𝐶 = 𝑅 𝑡 𝜏) = 12 (𝑒 − 160𝑢 44𝑢 ) 22000 = 14.4𝑢𝐴 Another Capacitor current problem • 𝐼𝐶 = 𝑡 −𝜏 𝑉𝑆 (𝑒 𝑅 ) = 𝐼𝑅 𝑉𝑅 = 𝑅 What is the current through a 1kΩ resistor with a .2uF capacitor connected in series with a after 280us if the source voltage is 18V? (Assume 0V for start up) − 𝑉𝑆 (𝑒 𝐼𝐶 = 𝑅 𝑡 𝜏) = 18 (𝑒 − 280𝑢 200𝑢 ) 1000 = 4.4𝑚𝐴 Voltage and current in LR circuits • Recall = 𝐿 (This is how long it takes to get 63.2% of max 𝑅 current through an inductor) • It turns out the equations for Voltage across an Inductor and Current through an inductor are as follows: 𝑉𝑆 (1 − 𝑒 𝐼𝐿 = 𝑅 − 𝑡 𝜏) 𝑉𝐿 = 𝑡 − 𝑉𝑆 (𝑒 𝜏 ) Inductor Problems 𝑡 𝑉𝑆 (1 − 𝑒 −𝜏 ) 𝐼𝐿 = 𝑅 𝑉𝐿 = 𝑉𝑆 (𝑒 − 𝑡 𝜏) • An 8H inductor and 1kOhm resistor are connected in series to a 10V source. Calculate the inductor current at t = 6ms. • Calculate the inductor voltage at this same time. • Calculate the resistor voltage at this same time. • Calculate the resistor current at this time. Another Inductor Problem 𝑡 𝑉𝑆 (1 − 𝑒 −𝜏 ) 𝐼𝐿 = 𝑅 𝑉𝐿 = 𝑉𝑆 (𝑒 − 𝑡 𝜏) • Calculate the inductor current at t = 5us after the switch is turned on for a 5mH inductor and a 2.2kOhm series resistance if the source voltage is 24V. • Calculate the inductor voltage at this same time. • Calculate the resistor voltage at this same time.
