Assignment 13: RC Circuits

Due: 8:00am on Wednesday, February 22, 2012

Note: To understand how points are awarded, read your instructor's Grading Policy.

Throw the Switch

In this problem denotes the emf provided by the source, and is the resistance of each bulb.

Part A

Bulbs A, B, and C in the figure are identical and the switch is an ideal

conductor. How does closing the switch in the figure affect the potential difference?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Find the potential difference across bulb C when the switch is closed

Hint not displayed

Hint A.3 Find the potential difference across bulb B when the switch is closed

Hint not displayed

Hint A.4 Find the potential difference across bulb A when the switch is closed

Hint not displayed

Hint A.5 Find the potential difference across bulb A when the switch is open

Hint not displayed

Check all that apply.

ANSWER:

The potential difference across A is unchanged.

The potential difference across B drops to zero.

The potential difference across A increases by 50%.

The potential difference across B drops by 50%.

Correct

Every time the ends of a resistor are joined together, or connected through an ideal conductor, the voltage across

the resistor drops to zero and the resistor is said to be short-circuited.

Part B

One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown in the

figure. Bulbs A, B, C, and D are identical and the switch is an ideal

conductor. How does closing the switch in the figure affect the potential difference?

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Find the equivalent resistance of the circuit when the switch is closed

Hint not displayed

Hint B.3 Find the voltage across bulb A when the switch is closed

Hint not displayed

Hint B.4 How to determine whether choice D is correct

Hint not displayed

Hint B.5 Find the voltage across bulb B when the switch is closed

Hint not displayed

Hint B.6 Find the voltage across bulb B when the switch is open

Hint not displayed

Check all that apply.

ANSWER:

The potential difference across A increases.

The potential difference across B doubles.

The potential difference across B drops to zero.

The potential difference across D is unchanged.

Correct

RC Circuit and Current Conceptual Question

In the diagram below, the two resistors, and , are identical and the

capacitor is initially uncharged with the switch open.

Part A

How does the current through compare with the current through immediately after the switch is first closed?

Hint A.1 Using Kirchhoff's junction rule for currents

Hint not displayed

ANSWER: The current through is greater than Correct the current through .

Part B

How does the current through compare with the current through a very long time after the switch has been

closed?

Hint B.1 Using Kirchhoff's junction rule for currents

Hint not displayed

Hint B.2 Current associated with a fully charged capacitor

Hint not displayed

ANSWER: The current through is equal to Correct the current through .

Part C

How does the current through compare with the current through immediately after the switch is opened (after

being closed a very long time)?

Hint C.1 Effect of a discharging capacitor

Hint not displayed

ANSWER: The current through is less than Correct the current through .

± Charging and Discharging a Capacitor in an R-C Circuit

Learning Goal: To understand the dynamics of a series R-C circuit.

Consider a series circuit containing a resistor of resistance and a capacitor of capacitance connected to a source

of EMF with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the

switch is open and the capacitor discharged.

Let us try to understand the processes that take place after the switch is closed. The charge of the capacitor, the

current in the circuit, and, correspondingly, the voltages across the resistor and the capacitor, will be changing.

Note that at any moment in time during the life of our circuit, Kirchhoff's loop rule holds and indeed, it is helpful:

, where is the voltage across the resistor, and is the voltage across the capacitor.

Part A

Immediately after the switch is closed, what is the voltage across the capacitor?

ANSWER:

zero

Correct

Part B

Immediately after the switch is closed, what is the voltage across the resistor?

ANSWER:

zero

Correct

Part C

Immediately after the switch is closed, what is the direction of the current in the circuit?

ANSWER:

clockwise

counterclockwise

There is no current because the capacitor does not allow the current to pass through.

Correct

While no charge can physically pass through the gap between the capacitor plates, it can flow in the rest of the

circuit. The current in the capacitor can be thought of as a different sort of current, not involved with the flow of

charge, but with an electric field that is increasing with time. This current is called the displacement current. You

will learn more about this later. Of course, when the charge of the capacitor is not changing, then there is no

current.

Part D

After the switch is closed, which plate of the capacitor eventually becomes positively charged?

ANSWER:

the top plate

the bottom plate

both plates

neither plate because electrons are negatively charged

Correct

Part E

Eventually, the process approaches a steady state. In that steady state, the charge of the capacitor is not changing.

What is the current in the circuit in the steady state?

Hint E.1 Charge and current

Hint not displayed

ANSWER:

zero

Correct

Part F

In the steady state, what is the charge of the capacitor?

Hint F.1 Voltage in the steady state

Hint not displayed

Express your answer in terms of any or all of , , and .

ANSWER:

=

Correct

Part G

How much work is done by the voltage source by the time the steady state is reached?

Hint G.1 Charge and EMF

Hint not displayed

Express your answer in terms any or all of , , and .

ANSWER:

=

Correct

In order to charge the capacitor, a total amount of charge had to move across the potential difference of

the EMF source. The source did work to move this charge equal to . Recall that a charged capacitor

stores an amount of energy . This is only half the work done by the EMF source. The remaining was

dissipated in the resistor. So such a simple charging circuit has a high loss percentage, independent of the value

of the resistance of the circuit.

Even though energy is dissipated across the resistor as the capacitor charges, note that the work done depends on

, but not on ! This is because it is the capacitor that determines the amount of charge flow through the circuit.

Charge flow stops when . The resistance does however affect the rate of charge flow i.e. the current. You

will calculate this effect in the parts that follow.

Now that we have a feel for the state of the circuit in its steady state, let us obtain expressions for the charge of the

capacitor and the current in the resistor as functions of time. We start with the loop rule: . Note that

, , and . Using these equations, we obtain , and then, .

Part H

Integrate both sides of the equation to obtain an expression for .

Hint H.1 Constant of integration

Hint not displayed

Express your answer in terms of any or all of , , , and . Enter exp(x) for .

ANSWER:

= Correct

Part I

Now differentiate to obtain an expression for the current .

Express your answer in terms of any or all of , , , and . Enter exp(x) for .

ANSWER:

= Correct

Theoretically, the steady state is never reached: The exponential functions approach their limits as

asymptotically. However, it does not take very long for the values of and to get very close to their limiting

values. The next few questions illustrate this point. Note that the quantity has dimensions of time and is called

the time constant, or the relaxation time. It is often denoted by . Using , one can rewrite the expressions for

charge and current as follows:

and

.

Graphs of these functions are shown in the figure.

Part J

Find the time that it would take the charge of the capacitor to reach 99.99% of its maximum value given that

and .

Hint J.1 Find an expression for the time

Hint not displayed

Express your answer numerically in seconds. Use three significant figures in your answer.

ANSWER:

=

5.53×10−2

Correct

Notice how quickly the circuit approaches steady state for these typical values of resistance and capacitance!

Let us now consider a different R-C circuit. This time, the capacitor is initially charged ( ), and there is no

source of EMF in the circuit. We will assume that the top plate of the

capacitor initially holds positive charge. For this circuit, Kirchhoff's loop rule gives , or equivalently,

.

Part K

Find the current as a function of time for this circuit.

Hint K.1 Find the charge on the capacitor

Hint not displayed

Express your answer in terms of , , , and . Enter exp(x) for .

ANSWER:

= Correct

The negative value of the current can be explained by the fact that the positive charge on the capacitor's top plate

decreases. Graphs of these functions are shown in the figure.

Exercise 26.40

A 5.00 capacitor that is initially uncharged is connected in series with a 5.10 resistor and an emf source with

180 negligible internal resistance.

Part A

Just after the circuit is completed, what is the voltage drop across the capacitor?

ANSWER:

=

0

Correct

Part B

Just after the circuit is completed, what is the voltage drop across the resistor?

ANSWER: = 180

Correct

Part C

Just after the circuit is completed, what is the charge on the capacitor?

ANSWER:

=

0

Correct

Part D

Just after the circuit is completed, what is the current through the resistor?

ANSWER:

=

3.53×10−2

Correct

Part E

A long time after the circuit is completed (after many time constants) what is the voltage drop across the

capacitor?

ANSWER:

=

180

Correct

Part F

A long time after the circuit is completed (after many time constants) what is the voltage drop across the resistor?

ANSWER:

=

0

Correct

Part G

A long time after the circuit is completed (after many time constants) what is the charge on the capacitor?

ANSWER:

=

9.00×10−4

Correct

Part H

A long time after the circuit is completed (after many time constants) what is the current through the resistor?

ANSWER:

=

0

Correct

Changing Capacitance Yields a Current

Each plate of a parallel-plate capacator is a square with side length , and the plates are separated by a distance .

The capacitor is connected to a source of voltage . A plastic slab of thickness and dielectric constant is

inserted slowly between the plates over the time period until the slab is squarely between the plates. While the

slab is being inserted, a current runs through the battery/capacitor circuit.

Part A

Assuming that the dielectric is inserted at a constant rate, find the current as the slab is inserted.

Hint A.1 What is the effect of the dielectric on capacitance?

Hint not displayed

Hint A.2 What is the current in the circuit?

Hint not displayed

Hint A.3 What is the initial capacitance?

Hint not displayed

Hint A.4 What is the change in capacitance?

Hint not displayed

Express your answer in terms of any or all of the given variables , , , , , and , the permittivity of free

space.

ANSWER:

=

Correct

Exercise 26.48

A 14.0 capacitor is charged to a potential of 50.0 and then discharged through a 170 resistor.

Part A

How long does it take the capacitor to lose half of its charge?

ANSWER:

=

1.65

Correct

Part B

How long does it take the capacitor to lose half of its stored energy?

ANSWER:

=

0.825

Correct

Problem 26.82

A capacitor that is initially uncharged is connected in series with a resistor and an emf source with

and negligible internal resistance.

Part A

Just after the connection is made, what is the rate at which electrical energy is being dissipated in the resistor?

ANSWER:

=

2460

Correct

Part B

What is the rate at which the electrical energy stored in the capacitor is increasing?

ANSWER:

=

0

Correct

Part C

What is the electrical power output of the source?

ANSWER:

=

2460

Correct

Part D

At a long time after the connection is made, what is the rate at which electrical energy is being dissipated in the

resistor?

ANSWER:

=

0

Correct

Part E

What is the rate at which the electrical energy stored in the capacitor is increasing?

ANSWER:

=

0

Correct

Part F

What is the electrical power output of the source?

ANSWER:

=

0

Correct

Part G

At the instant when the charge on the capacitor is one-half its final value, what is the rate at which electrical

energy is being dissipated in the resistor?

ANSWER:

=

614

Correct

Part H

What is the rate at which the electrical energy stored in the capacitor is increasing?

ANSWER:

=

614

Correct

Part I

What is the electrical power output of the source?

ANSWER:

=

1230

Correct

Problem 26.84

A resistor with 840 is connected to the plates of a charged capacitor with capacitance 4.52 . Just before the

connection is made, the charge on the capacitor is 6.80 .

Part A

What is the energy initially stored in the capacitor?

ANSWER:

=

5.12

Correct

Part B

What is the electrical power dissipated in the resistor just after the connection is made?

ANSWER:

=

2690

Correct

Part C

What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has

decreased to half the value calculated in part A?

ANSWER:

=

1350

Correct