Problem Solving 8: RC and LC Circuits Solutions

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Problem Solving 8: RC and LC Circuits Solutions OBJECTIVES. To understand how the charge on a capacitor and the current through a charging RC circuit behave as a function of time.. To derive and solve the differential equation for the charge on a capacitor in a charging RC circuit. 3. Explain the meaning of the time constant for the current in a charging RC circuit. 4. To understand how to recognize that an undriven LC circuit is a simple harmonic oscillator. 5. To derive and solve the differential equation associated with LC undriven circuits in order to find expressions for charge and current as a function of time. 6. To know how to apply conservation of energy to undriven LC circuits to determine charge and current at a given instant. Problem Charging a Capacitor Consider the circuit shown below. The capacitor is connected to a DC source of emf!. At time t = 0, the switch S is closed. The capacitor initially is uncharged, Q(t = 0) = 0. Question : At t = 0, what is the current in the circuit? In particular for t < 0, there is no electric potential difference across the capacitor so the capacitor acts like a short circuit. At t = 0, the switch is closed and current begins to flow according to! I 0 = R

2 At this instant, the potential difference from the battery terminals is the same as that across the resistor. This initiates the charging of the capacitor. As the capacitor starts to charge, the electric potential difference across the capacitor increases in time. The electric potential differences across a capacitor are summarized in the figure below. Question : Using Kirchhoff s loop rule, find the differential equation satisfied by the charge Q(t) on the capacitor. We choose ±Q on the capacitor, assign a direction to the current I, and traverse the loop in a clockwise direction as shown in the figure below. Using Kirchhoff s loop rule that the sum of the potential differences across each element is zero we find that 0 =! " I R " Q C According to our choices the current and charge are related by I = +dq / dt, therefore the differential equation satisfied by the charge Q(t) on the capacitor is given by 0 =! " dq dt R " Q C Question 3: Using your differential equation that you found in Question, explain in your own words how that the charge on the capacitor behaves as a function of time.

3 Since I must be the same in all parts of the series circuit, the current across the resistance R is equal to the rate of increase of charge on the capacitor plates. The current through the circuit will continue to decrease because the charge already present on the capacitor makes it harder to put more charge on the capacitor. Once the charge on the capacitor plates reaches its maximum value, the current in the circuit will drop to zero. This is evident by rewriting the loop law as I(t) R =! " Q C We shall solve the differential equation dq dt R =! " Q you found in Question by the method C of separation of variables. The first step is to separate terms involving charge and time, (this means putting terms involving dq and Q on one side of the equality sign and terms involving dt on the other side), dq #! " Q & $ % C ' ( = R dt ) dq Q " C! = " RC dt Question 4: Integrate both sides of the above equation to find an expression for the charge Q(t) on the capacitor as a function of time. You will be setting up definite integrals with limits of integration that cover the time interval [0,t] and the corresponding charge interval [0,Q(t)]. We can integrate both sides of the above equation, which yields Q(t) d Q! $ = " 0! RC Q " C# # Q(t)! C" & ln $ %!C" ' ( =! t RC Question 5: If you haven t already done so you can now exponentiate both sides of your result from Question 4 using the fact that exp(ln x) = x to yield to find an expression for Q(t). Exponentiating both sides yields $ 0 t d t! 3

4 A little algebraic rearrangement yields # Q(t)! C" & $ %!C" ' ( = e!t / RC. Q(t) = C!(" e "t / R C ) Question 6: What is the maximum value of the charge on the capacitor? In the limit as time approaches infinity, lim e #t / R C = 0, and so the maximum charge is t!" Q max = C!. Question 7: Make a plot of the charge Q(t) as a function of time t. Label all appropriate values on your plot. The time dependence of Q(t) is plotted in the below: Question 8: After a very long time, how does the electric potential difference across the capacitor compare to the electric potential difference across the battery? We know the charge on the capacitor we also can determine the electric potential difference across the capacitor,!v C (t) = Q(t) C = "(# e#t / R C ) 4

5 After a sufficiently long time the charge on the capacitor approaches the value Q(t =!) = C" = Q max. At that time, the electric potential difference across the capacitor is equal to the applied electric potential difference across the battery and the charging process effectively ends,!v C = Q(t = ") C = Q max C = # Question 9: Find an expression for the current I(t) through the circuit as function of time t. The current is equal to the derivative in time of the charge, I(t) = dq dt =! R e"t / R C = I 0 e "t / R C. The coefficient in front of the exponential is equal to the initial current I 0 =! / R through the circuit when the switch was closed at t = 0. Question 0: Make a plot of the current I(t) as a function of time t. Label all appropriate values on your plot. The graph of current as a function of time is shown in the figure below: The current in the charging circuit decreases exponentially in time, I( t) I e!t/ RC = 0. This t/ function is often written as I( t) = I0 e "! where! = RC is called the time constant. The time constant! is a measure of the decay time for the exponential function. This decay rate satisfies the following property: I( t +! ) = I( t) e " 5

6 which shows that after one time constant! has elapsed, the current falls off by a factor of e! = 0.368, as indicated in the figure above. Similarly, the electric potential difference across the capacitor (see figure below) can also be expressed in terms of the time constant! :!V C (t) = "(# e #t /$ ). Notice that initially at time t = 0, V ( t = 0) = 0. After one time constant! has elapsed, the C potential difference across the capacitor plates has increased by a factor (! e! ) = 0.63 of its final value:!v C (" ) = #($ e $ ) = 0.63#. 6

7 Problem : LC Circuit Consider a circuit consisting of a coil with inductance L, a resistor with resistance R, a capacitor with capacitance C, an applied dc voltage source,!, and two switches S and S connected as shown in the figure below. Question : Switch S is closed and switch S is left open. After a very long time interval time!t >> RC what is the charge on the capacitor? Answer: After a very long time, the charge on the capacitor is fully charged and the charge on the capacitor is Q f = C!. Now reset your clock so that at t = 0 switch S is then opened and switch S is closed, disconnecting the voltage source, leaving the coil and the charged capacitor connected together. Question : At t = 0, what are the initial values of the current I(t = 0) and charge on the capacitor Q(t = 0)? Answer: At t = 0, I(t = 0) = 0 and Q(t = 0)! Q 0 = C". Question 3: What differential equation does the charge Q(t) on the capacitor satisfy? Answer: The differential equation associated with the circuit is 7

8 Q di =! L. C dt The charge on the capacitor is related to the current according to So the circuit equation becomes dq = I. dt d Q dt + Q LC = 0. This equation is called the simple harmonic oscillator (SHO). A general form for the solution to the SHO equation is given by the sinusoidally oscillating function Q(t) = Acos(! 0 t + ") where! 0 is the angular frequency,! is a phase constant and A is the amplitude determined by the initial conditions. Question 4: What are! and A? Answer: The current is then the derivative of the charge with respect to time and is given by I(t) = dq(t) dt At t = 0, the charge and current are then =!" 0 Asin(" 0 t + #). Q 0 = Acos(!) = C" I 0 =!" 0 Asin(#) = 0. From the current initial condition we see that! = 0, and because cos(0) =, A = C!. Question 5: What is the angular frequency and period of oscillation of the charge on the capacitor? Answer: From our values for! and A, we have that the charge and current are given by 8

9 Q(t) = C! cos(" 0 t) I(t) = dq(t) dt =!" 0 C# sin(" 0 t) In order to find! 0, we take the second derivative of the charge and substitute this back into the SHO equation. Then becomes di dt = d Q =!" dt 0 C# cos(" 0 t) d Q dt + Q LC = 0!" 0 C# cos(" 0 t) + LC C# cos(" 0 t) = 0. This last equation is satisfied when! 0 = / LC. The period of oscillation is defined to be T! " # 0 = " LC. Question 6: Based on your answers to Questions 4 and 5, what are the expressions for charge on the capacitor and current in the circuit as functions of time? Answer: Combining our results, the expressions for the charge on the capacitor and current in the circuit as functions of time are given by Q(t) = C! cos( / LC t) I(t) = dq(t) =! C" sin / LC t dt LC ( ) 9

10 Question 7: A plot of Q(t) vs. t with! = 4 V, C! = nc and T = µs. What is the value of the self-inductance L? Answer: The capacitance is C = nc/4 V = 0.5! 0 "9 F. The self-inductance is L = T 4! C = ( " 0 #6 s) 4! (0.5 " 0 #9 F) =.0 " 0#4 H Question 8: What is the current in the circuit when the magnetic energy stored in the coil is exactly equal to the electric energy stored in the capacitor? Use the values given in Question 7 and your answer to Question 7. Answer: The initial stored energy is all in the electric field and equal to U 0 = Q C = C!. At an arbitrary time the energy in the circuit is given by U = U elec + U mag = If we consider the instant when U elec = U mag then Q C + LI. U = U elec + U mag = U mag = LI. Because there is no resistance, the energy is constant, hence U 0 = U. Therefore when the magnetic energy stored in the coil is exactly equal to the electric energy stored in the capacitor, 0

11 C! = LI. We can now solve for the current I = C L! = (0.5 " 0 #9 F) (.0 " 0 #4 H) (4 V) = 4.4 " 0#3 A