CHARGE BUILDUP AND DECAY IN CAPACITORS
RC CIRCUITS
In the next section you will measure what happens to the voltage across a charged capacitor when it is placed in series
with a resistor in a direct current circuit. Before making these measurements you should make some qualitative
observations of capacitor behavior, so that you can explain what is happening as charge decays off a capacitor. For the
observations in this section, you will need:
• Grey Power supply
• 1 #14 bulb
• 1 #48 bulb
• 2 capacitors, .47 F
• 1 multimeter
• 6 alligator clip wires
Qualitative Observations
By using a flashlight bulb as a resistor and one or more of the amazing new capacitors
that have capacitances up to about a farad in a tiny container, you can “see” what
happens to the current flowing through a resistor (the bulb) when a capacitor is charged
by a battery and when it is discharged.
. A #14 bulb
(rounded)
Activity: Capacitors, Batteries, and Bulbs
a. Connect a rounded #14 bulb in series with the 0.47 F capacitor, a switch, and the 4.5 V power supply. Describe
what happens when you close the switch. Draw a circuit diagram of your setup.
b. Now, can you make the bulb light again without the battery in the circuit? Mess around and see what happens.
Describe your observations and draw a circuit diagram showing the setup when the bulb lights without a battery.
c. Draw a sketch of the approximate brightness of the bulb as a function of time when it is placed across a charged
capacitor without the battery present. Let t = 0 when the bulb is first placed in the circuit with the charged
capacitor. Note: Another way to examine the change in current is to wire an ammeter in series with the bulb.
d. Explain what is happening. Is there any evidence that charge is flowing between the “plates” of the capacitor as
it is charged by the battery with the resistor (the bulb) in the circuit, or as it discharges through the resistor? Is
there any evidence that charge is not flowing through the capacitor? Hints: (1) You may want to repeat the
observations described in a. and b. several times; placing the voltmeter across the capacitor or placing an
ammeter in series with the capacitor and bulb in the two circuits you have devised might aid you in your
observations. (2) Theoretically, how should the voltage across the capacitor be related to the magnitude of the
charge on each of its conductors at any given point in time?
e. What happens when more capacitance is put in the circuit? When more resistance is put in the circuit? (You
can use a #48 bulb—the oblong one—in the circuit to get more resistance.) Hint: Be careful how you wire the
extra capacitance and resistance in the circuit. Does more capacitance result when
capacitors are wired in parallel or in series? How should you wire resistors to get more
resistance?
. A #48 bulb
(elongated).
A Capacitance Puzzle
Suppose two identical .47 F capacitors are hooked up to 3.0 V and 4.5 V batteries in two separate circuits. What would
the final voltage across them be if they were each unhooked from their batteries and hooked to each other without being
discharged? This situation is shown below. Warning: Be sure to hook the terminals that were charged positively
together and those that were charged negatively together.
A capacitor circuit. What is the multimeter reading Vf?
Activity: Proof of the Puzzler
a. What do you predict will happen to the voltage across the two capacitors? Why?
b. Can you use equations to calculate what might happen? Hint: What do you know about the initial charge on
each capacitor? What do you know about the final sum of the charges on the two capacitors, if there is no
discharge?
c. Set up the circuit and describe what actually happens.
d. How well did your prediction hold? Explain.
QUANTITATIVE MEASUREMENTS ON AN RC SYSTEM
The next task is to do a more quantitative study of your “RC” system. We will do this in two ways.
The first involves measuring the voltage across a charged capacitor as a function of time when a carbon resistor
has been placed in a circuit with it. A computer-based laboratory voltage logging setup can be used to obtain data and
view the trace of voltage vs. time in graphical form as the capacitor discharges. The goal here is to figure out the
mathematical relationship between voltage across the capacitor and time that best describes the voltage change as the
capacitor discharges.
For the activities in this section you will need:
•
•
•
•
•
•
•
•
•
•
1 power supply, 4.5 V
1 capacitor, approx. 5000 F
1 SPDT switch
1 resistor, 1.2 k
1 computer-based laboratory system
1 voltage measuring lead
1 voltage-logging software
1 digital multimeter
6 alligator clip wires
1 capacitance meter (for C  5000 F)
A bulb is not a good constant value resistor because its resistance is temperature dependent and rises when it is
heated up by current. For these more quantitative studies you should use a 1.0 k resistor in place of the bulb while
attempting to charge a 5000 F capacitor. Wire up the circuit shown below in Figure 5.10 with a two-position switch in
it. The switch will allow you to flip from a situation in which the battery is charging the capacitor rapidly to one in
which the capacitor is allowed to discharge through the resistor more slowly. The voltmeter and leads to your computer
interface should be placed in parallel with the capacitor—this allows you to measure the voltage across it.
RC circuit with voltage measurements across a discharging capacitor.
Activity: The Decrease of Voltage in an RC Circuit
a. Assume that the capacitor has been charged by the battery. What do you predict will happen to the voltage
across the capacitor when the two-position switch is flipped so that the battery is removed from the circuit?
Explain the reasons for your prediction.
b. As soon as the switch is flipped from a battery terminal to a terminal of the capacitor, you can start measuring
the voltage across the capacitor at least every one or two seconds until the voltage across the capacitor is about
1/100th of its initial value. Graph your data and guess how to model it. Affix (1) a table summarizing your data;
(2) a graph of the V vs. t data and model in the following space. Note: Be sure to save your data on disk, as you
will need to use it again in Activity 5.7.3. Hint: The V vs. t curve cannot be modeled using a simple power or
-kt
root of V. Thus, you should try some simple functions to model V such as: k/t, k, e , and so on where the k is a
constant.
c. How did your observations fit with the prediction you made in part a?
The Theoretical RC Decay Curve
If you made careful measurements of V vs. t for a capacitor, C, discharging through a resistor, R, you should have been
able to plot what is known as an exponential decay curve. Mathematical reasoning based on the application of Ohm’s
law as well as the definition of current and capacitance can be used to predict exponential decay given by
V  V0e
t
 RC
where V0 is the initial voltage across the capacitor and t represents the time elapsed since the switch was thrown to cut
the battery out of the circuit.
Activity: Derivation of the Theoretical Decay Curve
a. What is the equation for the voltage difference across the capacitor shown below? Express this in terms of the
charge qc on the capacitor and C? Hint: What is the definition of C?
b. What is the equation for the voltage difference across the resistor in terms of the current, I, flowing through it
(due to the discharge of the capacitor) and its resistance, R? Hint: Recall that voltage drops in the direction of
current flow.
c. Assume that the switch is in the position shown in Figure 5.11 and that qc represents the charge on the capacitor.
Show that at all times while the capacitor is discharging,
qc
 IR
C
Note: As the capacitor discharges, qc gets smaller and smaller.
d. Using the definition of the instantaneous electric current passing through the resistor, explain why
I 
dqc
dt
where qc represents the charge on the capacitor (not the charge flowing through the resistor). Hints: (1) What is
the source of the charge flowing through the resistor? (2) What is the relationship between the rate of flow of
charge through the resistor, dqR/dt and the rate at which charge flows off the capacitor plates, dqc/dt?
e. Use the answers given previously to show that
dqc
q
 c
dt
RC
in the circuit under consideration.
f. Show that the equation qc = q0e
satisfies the condition that
(–t/RC)
, where q0 is a constant representing the initial charge on the capacitor,
dqc
q
 c
dt
RC
(–t/RC)
Hint: Take the derivative of qc with respect to t and replace q0e
with qc.
g. Use the definition of capacitance once again to show that theoretically we should expect that the voltage across
the capacitor will be given by
V = V0e
(–t/RC)
where V0 represents the initial voltage across the capacitor.
Transient RC Series Circuits
Charging a Capacitor:

The voltage across the capacitor is not instantaneously equal to that of the voltage across the battery when the
switch is closed. The voltage on the capacitor builds up as more and more charges flows onto the capacitor until
the battery is no longer able to "push" any more charge onto the capacitor, at which point the capacitor becomes
fully charged.

The initial flow of charges from the battery to the capacitor means that there is a current flowing through the
system until the capacitor is charged. This current flow decays exponentially from some initial value to zero.

The exponential nature of the change (of charge or voltage) with time results from applying Kirchoff's voltage
rule and the definition of current as the rate of change of charge with time. (See Next Section)

Note that in any real system the wires have some resistance, so a more correct model includes some resistance in
series with the capacitor before it is charged up (Again see Next Section on RC circuits).
The exponential nature of charging or discharging a
capacitor:
Mathematically, an exponential change occurs when the derivative of a
time is proportional to the quantity itself. For example if,
quantity with
Let us apply this to the discharge of a capacitor through a resistor when the switch is closed and the capacitor
is initially charged. After the switch is closed Kirchoff's voltage rule applies at all times and gives the
equation.
Using Ohm's Law V = IR, the definition Capacitance C= Q/V, and the definition of current I = dQ/dt, we can
rewrite this last equation so that the derivative of the charge is proportional to the negative of the amount of
charge on the capacitor.
Thus the charge on the capacitor decays exponentially with a time constant .
Instead of Qo we have use Qmax since the capacitor starts off with its maximum charge and decays away
exponentially.
Mathematically, discharging a capacitor takes an infinite amount of time.
The time constant  represent the time for the system to make significant change in charge, voltage,
or current whenever a capacitor is charging or discharging.
After a time equal to one time constant, t =, the charge on the capacitor has dropped to e-1 = 36.8%
of its maximum value. After 5 the charge has dropped to 0.7% of its maximum value.
t
e–
t/
 .368
2 .135
3 .050
4 .018
5 .007
Other quantities such as the current and the voltage drop across the resistor or the capacitor can be found using
the definition of current I = dQ/dt, Ohms Law V = IR, and the definition of capacitance C= Q/V.
Thus the current also decays away from its initial value when the switch is first closed. Ohms law shows that
this also true for the voltage across the resistor.
The mathematical results for charging a capacitor are similar but more complicated. The main difference is
that the charge and voltage across the capacitor approach a maximum value exponentially. For example, the
charge on the capacitor start at zero an goes to a maximum value,
Now comes the acid test: the comparison of the experimentally determined rate of the capacitor discharge to the
theoretically predicted rate.
Activity: Does the Observed Decay Curve Fit Theory?
a. Carefully measure the R and C of the resistor and capacitor you used in the experiment in the previous activity
using a digital multimeter and a capacitance meter. List these values below (with proper units, of course). Also
list the value of V0 from that experiment.
b. Develop a modeling worksheet. Set R, C, and V0 as absolute parameters, using your measured values from part
a, and calculate the theoretical V vs. t in Logger Pro using the Calculated Column tool for the circuit using the equations
you derived. Create an overlay plot of the theoretical and experimental values for V vs. t, using the experimental data
you obtained. Be sure to include copies of both for your blog.
c. How well does theory match with experiment in this case?
d. What do you think would happen to the decay time if R were doubled? If C were doubled?
MORE ON CHARGE DECAY IN CAPACITORS
A Qualitative Summary of RC Decay
Let’s consider the process of discharging a capacitor that is in series with a resistor one more time.
Activity: Explaining Discharging Qualitatively
a. Assume that the capacitor is fully charged. When the switch is first flipped
so the battery is no longer in the circuit, how much charge is on the
Fig. 5.11. Again!
capacitor C? What is the potential difference V0 across the plates?
b. Is the current through the resistor a maximum or a minimum just after the switch is flipped? Does the current
ever flow through the capacitor? Explain.
c. How is the potential difference across the resistor related to that across the capacitor?
d. What happens to the potential difference across the capacitor as charge drains away from it? Explain.
e. What happens to the potential difference across the resistor at the same time? Explain.
f. If the potential across the resistor starts to change, what must happen to the current in the circuit? Explain.
g. Why does the draining of charge from the capacitor eventually stop? Why does the current in the circuit go to
zero?
Capacitor Charging
If the resistor, R, in Figure 5.11 is moved up next to the battery as shown in Figure
5.12, an uncharged capacitor, C, can be charged by the battery in the presence of the
resistor. The qualitative and quantitative considerations of this situation are very
analogous to that of capacitor decay. For example, the capacitor charges up more
rapidly at first when there are no charges on either of the capacitor plates to repel each
other. Also, after a while when the voltage across the capacitor is equal to that across
Fig. 5.12.
the battery, the charging stops completely. It can be shown that the voltage, V, across
the capacitor as it is charging is described by the equation
V = Vb (1 - e
–t/RC
)
where Vb is the voltage across the battery. This charging equation is used in the design of the flashing lights used at road
construction sites and in the flash units used by photographers.
Transient RC Conditions:
Charging a Single Capacitor in a Series with a Resistor
Start
t=0
Some Time
Later at t
Long Time
Later t >> 
CAPACITOR
Voltage
0
VC,max = QC,max / C
Charge
0
QC,max
RESISTOR
Voltage
VR,max = IR,max R
0
Current
IR,max
0

Note that these equations can only be used in a complex RC circuit if the circuit can be reduced to a simple RC
circuit.

When the switch is first closed all the voltage is across the resistor and the circuit look like a simple DC Ohms
law circuit with a resistor and no capacitor. This condition gives the maximum value of the voltage across the
resistor and the maximum value of the current.

As charge flows onto the capacitor, the current drops exponentially (derivation) from its maximum value when
the switch is first closed, Imax = Vb / R and I(t) = Imax e-t/.

As charge flows onto the capacitor, the voltage across the resistor also drops exponentially. This happens
because the capacitor now has a voltage drop and the circuit conforms to Kirchoff voltage rule at any time, Vb =
Vc(t) + VR(t) - the voltage drop across the capacitor and the resistor must equal to the voltage of the battery.

After a long time (t >> ) the current stops flowing in the circuit as the capacitor becomes fully charged and its
voltage equals to that of the battery if it is a simple RC circuit. The voltage drop across the resistor is zero since
there is no current flowing and VR = I R.

The long time condition give the maximum charge on the capacitor. Here the circuit looks like a simple
capacitor circuit with no resistor, Qmax = CVb.

Note once again, that if the circuit is a more complex arrangement of capacitors and resistors them the maximum
values may not be the voltage of the battery but the maximum value of the voltage across that component.
Transient RC Conditions:
Discharging a Single Capacitor Across a Resistor
Start
t=0
Some Time Later
at some time t
Long Time Later
t >> 
CAPACITOR
Voltage
VC,max = QC,max / C
0
Charge
QC = QC,max
0
RESISTOR
Voltage
VR,max = IR,max R
0
Current
IR = IR,max
0

When the circuit is first closed. the capacitor acts as a source of charge which can flow through the resistor. The
capacitor acts like a emf source with a short lifetime compared to a battery.

In energy terms, the energy in the electric field stored in the capacitor is converted into the energy of current
flow which is in turn dissipated as heat as the charge flows through the resistor.

This is the simplest transient RC circuit problem since all the quantities that vary with time decay away
exponentially.
RC Circuit Problem
A 100 mF capacitor is connected in series with a 10.0  resistor. This combination is connected in parallel with a 25.0
 resistor. Both branches are then connected in parallel to a 4.50 V battery that can be switched on and off. The
capacitor starts off fully discharged.
(0) Draw a picture of the circuit
(A) What is the time constant in both branches when the switch is closed?
After the switch is closed and before it is opened we can break the problem into two loops and apply
Kirchhoff’s voltage rule to each loop.
Loop 1 is just that of a simple RC circuit when the capacitor is being charged up and has a time constant 1 = R 1 C
. Thus we can use the generic equations associated with charging a capacitor. The basic ones involve the current
through the resistor and the charge on the capacitor.
The voltage drop at any moment across either component can be found using Ohm’s Law and the
definition of capacitance.
The maximum value of the current can be found when the circuit is first switched on because all the
voltage drop of the battery is across the resistor at that moment. The maximum charge on the capacitor
can be found from knowing that when the capacitor is fully charged no current will flow and all the
voltage drop of the battery is across the capacitor.
In a more complex circuit, R1 and Q would have to be replaced by the effective resistance and effective
capacitance.
Loop 2 is just that of a simple DC circuit with resistor R2. Loop 2 does not have a time constant associated with
capacitance in the loop. The will reach is maximum value almost instantaneously and remain constant until the
switch is opened.
A) Find 1 when the switch is closed.
We start by breaking the circuit into two loops and apply Kirchhoff’s voltage rule to each loop. In doing
this we have reduce this problem into two problem. The first is just a simple RC circuit problem and the
second is just a simple Ohm’s law circuit.
Since there is only one resistor and one capacitor in loop 1, the time constant is very simple to find.
(B) What is the maximum charge that the capacitor can attain after the switch is closed?
Qmax can be found by looking at system after a long time, when t >> , and the capacitor is fully charged.
In this limit there is no current flowing through R1 so that there is no voltage drop across the resistor.
Since the resistor and the capacitor are in parallel with the battery, the voltage drop across the capacitor
must equal to that of the battery. Using the definition of capacitance,
Don’t be fooled into thinking that this is always true - that Qmax is always the voltage of the battery times
the capacitance.
(C) When will the voltage drop across the 10.0  resistor be equal to 1.50 V after the switch is closed?
We know that the current through R1 is exponentially decaying away from some maximum value since
loop 1 is equivalent to a simple RC circuit.
Since the capacitor has no charge initially, loop1 looks like a simple DC circuit problem at the start. As a
result, the maximum current flow through R1 occurs when the switch is first closed. More specifically,
the voltage drop across the capacitor is zero initially since it has no charge initially and VC = Q / C . Thus
initially the voltage of the battery is all across the resistor.
(D) If the switched is opened, what will be the value of the new time constant and in which direction will the current
flow through the 10.0  resistor?
When the switch is open the circuit flow changes to an almost new problem. The bottom branch with
the battery becomes an open circuit in which no current can flow. The top two branches become a
closed circuit powered by the charged stored in the capacitor - a discharging RC Circuit.
Since current flows from positive to negative, the current will flow from the positively charged side of
the capacitor around the circuit to the negative side of the capacitor. The positively charged side of the
capacitor is the side that was originally connected to the positive side of the battery. In this case the
current will flow counterclockwise around the circuit. Thus the current will flow from right to left
through the 10.0  resistor.
Without the capacitor in the circuit, the current in the middle branch would fall to zero almost
instantaneously. It will not fall to zero instantaneously because there is some very small inductance in
the wire and resistor in the middle branch. We will study this later when we look at LR circuits.
Since the top loop now contains two resistors that are in series with each other the effective resistance
of the circuit is just their sum. For this circuit the time constant is given by,
Note that this would be the time constant when the switch opened no matter how soon or how long the
circuit was closed initially. The time would effect how much charge has built up on the capacitor.
(E) If the switch is opened after the capacitor is fully charged, how long will it take for there to be only one electron
on the capacitor?
For a charged capacitor in a RC circuit the charge will decay away exponentially.
Note that time constant has now changed because there is another resistor in the circuit (see part D).
We found the maximum charge on the capacitor to be .450 C in part B. The charge of an electron is e =
1.602x10 –19 C.
In theory it should take an infinite amount of time for the capacitor
to
discharge because of the exponential nature of the decay. However, charge is quantized and once the
last electron has exited the capacitor it will be fully discharged.
In this problem the capacitor had approximately .450 C / 1.602x10 -19 C = 2.8x10 18 electrons initially.
Note that the time it would take to discharge 99.99% of its initial charge is 32.2 seconds which
represents about 9.2 time constants. This is why we say that the time constant represents the
characteristic time for the system to change.
(F) If the switch is opened after the capacitor is fully charged, how long will it take for the voltage across the 25.0 
resistor to drop to 1.5V?
We include this part of the problem to warn you of an easy trap to fall into. In the RC circuit topic the generic
equation for the voltage decay is,
Because there is only one resistor in the generic equations, Vmax equals to voltage of the charged
capacitor, which in our problem is 4.50 V.
However, there are two resistors in the circuit and the capacitor's voltage of 4.50 V will be divided up
between the two resistors. To solve this problem we need to find the voltage drop across the resistors at
moment the capacitor starts to discharge.
The initial current (which is also be the maximum current) can be found by treating the top loop as
though it we a DC circuit.
With this we can find the maximum voltage of R2.
Then we could use an equation like
to find the time for the voltage to drop across resistor R1 since the generic equation should really be
written as
What makes this confusing is if you were asked to find the time when the voltage across the 10.0 
resistor drop to 1.50 V when the switch was first closed then Vmax = 4.50 V because there is only one
resistor R1 in Loop 1 that is in parallel with the battery.
Observe that the voltage drop across R1 is 4.50 V before the switch is opened. When the switch is
opened, the middle bulb drops almost instantaneously to the same brightness as the bulb in the top
circuit and then decays away with a longer time constant of 2. The bulbs are equally as bright since
they have the nearly the same resistance and the voltage of the capacitor is divided between the two.
(In the actual demo it may not look that way if you focus on the enlarged images of the bulbs. This is
because movie camera automatically adjusts the brightness. The top bulb is dark when the switch is
opened while the middle bulb is shinning brightly when the switch is opened.)
Also note that when the switch is first closed at the start of the demo, the top branch bulb starts out
equally as bright as the middle branch bulb and decay away as the capacitor is being charged up.
Complex RC Circuit Problem
Given the circuit above:
(A) Determine the current through each of the resistors after the switch is closed and the capacitor is fully charged.
Relevant Physics:
This cannot be reduced to a simple RC circuit. However, we do know that when the capacitor is fully charge no
current will be flowing through the capacitor. In this situation we can apply Kirchhoff’s rules to analyze the
problem. The current flowing through resistor R1 and R3 will be the same (Kirchhoff’s current rule) as will the
current through resistors R2 and R4. This means we can treat the two branches as if they are effectively in parallel
with each other.
At the branch point 1, the potential is equal to that of the battery VB. If the potential drop across resistor
R1 is larger than the potential drop across resistor R2 then point “a” will have a lower potential that point
“b”.
To determine the value of these potential drops we will need to find the current in the two branches.
Here we can use Kirchhoff’s voltage rule or that the circuit can be reduced to an equivalent resistance in
parallel with the battery.
Once we determine the potential drops V1 = I1 R 1 and V2 = I2 R2 we can determine the potential at
points “ a” and “ b”,
Using the definition of capacitance we can now find the charge on the capacitor since the voltage drop
across the capacitor is |Va – V b|.
When the switch is opened , the capacitor will act like a “battery” supplying electrical charge until the
capacitor is drained. Note that the current will flow out of the capacitor in the opposite direction that the
current flowed into the capacitor when it was being charge up. The flow forms a new circuit that can be
reduced to a simple RC discharging circuit so that one can determine the time constant.
(A) Find the current through each of the resistors after the capacitor is fully charged.
This is not as big a task as it may seem at first because in this situation there is no current flowing through the
capacitor when it is fully charged. Knowing that the current flowing into point “a” must equal the current flowing
out (Kirchhoff’s current rule), the current that flows through R1 must also flow through R3. Call it I1. Likewise,
only one current will flowing through the bottom branch, call it I2.
With no current flowing through the capacitor both the top and bottom branchs are in parallel with the
battery. Thus (for each branch) we have two resistors in series with each other and together they are in
parallel with the battery. Thus the voltage drop across R1 and R3 is equal to that of the battery VB and
Similarly for the lower branch,
To check our results we note that .50 A + .90 A = 1.40 A must be the current flowing through (or
produced by) the battery. We can calculate this if we find the effective resistance of the two branches.
The effective resistance of the two branches is that of two parallel branches each containing two
resistors in series.
Our results check
(B) Determine the conditions under which the “a-side” of the capacitor will have a positive charge, i.e., which side is
at a higher potential after the switch is closed and the current attains steady state flow conditions. For the values
of the resistors given determine the potential on both sides of the capacitor at points “ a” and “ b”.
The potential of the positive side of the battery is +45 V. (By definition the potential of ground is zero. ) In fact
the potential is 45 V anywhere along the wires until it encounters a resistor (or a capacitor). This means that the
potential before the current inters either resistor R1 or R2 is 45 V. As the current goes through either resistor its
potential will drop by an amount equal to voltage found using Ohm’s Law, V = I R .
Since we already know the current in the two branches from part A.
and
It turns out that Vb > V a so it is really the bottom side of the capacitor which is positively charged. This
means that when the switch is first closed the current through the capacitor will flow from “b” to “a”
and after the switch is opened the current flow out of “b” through the circuit elements and back to “a”.
You could double check these results and see if the voltage drop across R3 is 25.0 V – 0 V = 25.0 V and
if the voltage drop across R4 is 27.0 V –0 V = 27.0 V, and they are.
It is possible to determine a very simple relationship among the resistors that would make Va > V b.
Symbolically, the potential drop across R1 and R2 are:
Thus the potential at the points “a” and “b” are,
The potential at “a” will be larger than the potential at “b” if,
For our values of the resistors we see that this indeed true.
(C) What is the maximum charge that the capacitor will attain after the switch is closed?
This part is now a snap since we found the potential at points “a” and “b” in part B. The absolute value of the
difference of the voltages is just the voltage drop across the capacitor.
(D) After Capacitor is fully charged and the switch is opened, how long will it take for the charge on the capacitor to
drop to half its maximum value?
After the battery is disconnected the circuit will look like,
This circuit can be reduced to that of a capacitor C and effective resistance Re,
The capacitor is discharging RC circuit so the charge is exponentially decaying,
We found Qmax in part C since it is equal to the final charge on the capacitor after the switch is first closed.
However, we will not need to actually know its value to solve this part of the question.
The time constant is given by,
When the charge is equal to half its initial value, we can solve charge decay equation for the time.
The time constant is quite short so it not surprising that our answer is also small.
What about the time constant in the first part of the problem when the capacitor is charging up? Since we
cannot reduce this part to a simple RC circuit we would have to solve a set of six simultaneous differential
equations - each branch will have its own current flow and time constant if all the resistors have different
values. Here we would have to apply Kirchhoff’s rules to generate the six differential equations. The solution
of these equations is beyond the scope of this course.