MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Problem Solving 8: RC and LC Circuits Solutions
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 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 1 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 .
1
Question 1: At t  0 , what is the current in the circuit?
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
We shall solve the differential equation
dQ
Q
R
dt
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 d Q and Q on one side of
the equality sign and terms involving dt on the other side),
dQ
1
 dt

Q R
   C 

dQ
1

dt
Q  C
RC
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
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
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 0 e  t / RC . This
function is often written as I (t )  I 0 e t / where   R C 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 1 ,
which shows that after one time constant  has elapsed, the current falls off by a factor of
e 1  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  :
VC (t)   (1 et /  ) .
5
Notice that initially at time t  0 , VC (t  0)  0 . After one time constant  has elapsed, the
potential difference across the capacitor plates has increased by a factor (1  e1 )  0.632 of its
final value:
VC ( )   (1 e1 )  0.632  .
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Problem 2: 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 S1 and S 2
connected as shown in the figure below.
Question 1: Switch S1 is closed and switch S 2 is left open. After a very long time interval time
t  RC what is the charge on the capacitor?
Answer:
Now reset your clock so that at t  0 switch S1 is then opened and switch S 2 is closed,
disconnecting the voltage source, leaving the coil and the charged capacitor connected together.
Question 2: At t  0 , what are the initial values of the current I(t  0) and charge on the
capacitor Q(t  0) ?
Answer:
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Question 3: What differential equation does the charge Q(t) on the capacitor satisfy?
Answer:
A general form for the solution to the SHO equation is given by the sinusoidally oscillating
function
Q(t)  Acos( 0t   )
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:
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Question 5: What is the angular frequency and period of oscillation of the charge on the
capacitor?
Answer:
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:
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Question 7: A plot of Q(t) vs. t with   4 V , C  2 nC and T  2 s . What is the value of
the self-inductance L ?
Answer:
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:
10