Problem Solving 8: RC and LR Circuits


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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Problem Solving 8: RC and LR Circuits Section Table and Group (e.g. L04 3C ) Names Hand in one copy per group at the end of the Friday Problem Solving Session. OBJECTIVES 1. To understand how the charge on a capacitor and the current through a charging RC circuit behave as a function of time. 2. 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 the current through an inductor in an LR circuit behaves as a function of time for the case when a switch in the circuit is opened. 5. To derive and solve the differential equation for the current through an inductor in an LR circuit when a switch is opened. Problem 1 Charging a Capacitor Consider the circuit shown below. The capacitor is connected to a DC source of emf ε. At time t = 0, the switch S is closed. The capacitor initially is uncharged, Q(t = 0) = 0. Question 1: At t = 0, what is the current in the circuit? 1
2 Solution: 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 2: Using Kirchhoff s loop rule, find the differential equation satisfied by the charge Q(t) on the capacitor. Question 3: Using your differential equation that you found in Question 2, explain in your own words how that the charge on the capacitor behaves as a function of time. 2
3 We shall solve the differential equation dq dt R = ε Q C you found in Question 2 by the method 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 ε Q C = 1 R dt dq Q Cε = 1 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)]. 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). 3
4 Question 6: What is the maximum value of the charge on the capacitor? Question 7: Make a plot of the charge Q(t) as a function of time t. Label all appropriate values on your plot. 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? 4
5 Question 9: Find an expression for the current I(t) through the circuit as function of time t. Question 10: Make a plot of the current I(t) as a function of time t. Label all appropriate values on your plot. 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 τ where τ = RC is called the time constant. The time constant τ is a measure of the decay time for the exponential function. This decay rate satisfies the following property: 1 I( t + τ ) = I( t) e, which shows that after one time constant τ has elapsed, the current falls off by a factor of 1 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 τ : ΔV C (t) = ε(1 e t /τ ). 5
6 Notice that initially at time t = 0, ΔV C (t = 0) = 0. After one time constant τ has elapsed, the potential difference across the capacitor plates has increased by a factor final value: ΔV C (τ ) = ε(1 e 1 ) = 0.632ε. 1 (1 e ) = of its 6
7 Problem 2: Opening a Switch on an RL Circuit The LR circuit shown in the figure below contains a resistor R 1 and an inductance L in series with a battery of emf ε 0. The switch S is initially closed. At t = 0, the switch S is opened, so that an additional very large resistance R 2 (with R 2 >> R 1 ) is now in series with the other elements. Question 1: If the switch has been closed for a long time before t = 0, what is the steady current I 0 in the circuit? Question 2: While this current I 0 is flowing, at time t = 0, the switch S is opened. Write the differential equation for I(t) that describes the behavior of the circuit at times t 0. 7
8 Question 3: Solve this equation (by integration) for I(t) under the approximation that ε 0 = 0. (Assume that the battery emf is negligible compared to the total emf around the circuit for times just after the switch is opened.) Express your answer in terms of the initial current I 0, and R 1, R 2, and L. 8
9 Question 4: Using your results from part b), find the value of the total emf around the circuit (which from Faraday's law is LdI / dt ) just after the switch is opened. Question 5: How reasonable is your assumption in part b) that ε 0 could be ignored for times just after the switch is opened? Question 6: What is the magnitude of the potential drop across the resistor R 2 at times t > 0, just after the switch is opened? Express your answers in terms of ε 0, R 1, and R 2. 9
10 Question 7: How does the potential drop across R 2 just after t = 0 compare to the battery emf ε 0, if R 2 = 80R 1? Question 8: Based on your result from Question 7, why should you be very careful when you open the switch? 10
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