This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PS11 Solutions-1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 11 Solutions Problem 1: Where’s the energy go? Consider a simple RC circuit consisting of a “variable” battery (EMF 0 V or ε ), a resistor R and an uncharged capacitor C all in series. At time t=0 the battery is “turned on” (EMF ε ) and the capacitor is allowed to charge. A time T (>> RC ) later the battery is turned off (EMF 0 V) and the capacitor discharges. This process is continually repeated. (a) How much energy does the resistor dissipate while the capacitor is being charged (i.e. between t=0 and t= T )? The resistor dissipates power P = I 2 R . To find the energy dissipated we need to integrate: ( ) ( ) N 2 2 2 2 2 2 2 2 2 1 2 1 2 2 T T T t t t t T T t t T t U P t dt I t Rdt e Rdt R e e d t e C R R R τ τ τ τ ε ε ε τ ε τ ε − = = = − − − ≈ = ⎛ ⎞ ∆ = = = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = = − = − ≈ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ∫ ∫ ∫ ∫ (b) How does this compare to the amount of energy stored in the capacitor at time T ? They are equal. (c) How much energy does the resistor dissipate while the capacitor is discharging (i.e. between t= T and t=2 T )? Since it gets rid of all the energy in the capacitor during this time, it’s the same: 2 1 2 C ε (d) What is the time average power dissipated by the resistor? Hint: To average a set of numbers you add the numbers and divide by the number you added. To average a periodic function, integrate it over one period and divide by the length (or in this case time ) of the period. We already did the integral. It dissipated 2 1 2 C ε every T so 2 1 2 T P C ε = PS11 Solutions-2 Problem 2: LC Circuit (a) Initially, the capacitor in a series LC circuit is charged. A switch is closed, allowing the capacitor to discharge, and after time T the energy stored in the capacitor is one- quarter its initial value. Determine L if C and T are known. If the energy is at ¼ of it’s initial value then the voltage has fallen in half. That is, ( ) ( ) 2 2 1 9 cos 2 3 3 V T V T V T T L T C LC π π ω ω ω π = = ⇒ = ⇒ = = ⇒ = (b) A capacitor in a series LC circuit has an initial charge Q and is being discharged. Find, in terms of L and C , the flux through each of the N turns in the coil at time t , when the charge on the capacitor is Q ( t ). The flux depends on the current and inductance: L I N = Φ . So we just need to write the current in terms of the charge. By conservation of energy 2 2 2 1 2 2 2 Q C L I Q C + = . So, ( ) ( ) 2 2 Q Q t LI L N N L C − Φ = = (c) An LC circuit consists of a 10.0-mH inductor and a 0.100- F µ capacitor. If the maximum instantaneous current is 0.200 A, what is the greatest potential difference across the capacitor?...
View Full Document
This note was uploaded on 02/22/2009 for the course 8 02 taught by Professor Sciolla during the Fall '08 term at MIT.
- Fall '08