Oct. 619.1

Oct. 819.2-19.4

Oct. 10boardwork

Quiz 4

Oct. 1319.5-19.9

Oct. 15boardwork

Oct. 17boardwork

Quiz 5

Oct. 20review

Oct. 22Exam 2

Ch. 18, 19

Oct. 2420.1-20.2

Upcoming Schedule

“Physics is becoming too difficult for the physicists.”—mathematician David Hilbert, referring to quantum mechanics

19.5 EMF’s in Series and in Parallel: Charging a Battery

If you put batteries in series the “right way,” their voltages add:

+ =6 V 3 V 9 V

+ =6 V 3 V 3 V

If you put batteries in series the “wrong way,” their voltages add algebraically:

magnitudes only

chosen loop direction -6 V +3 V -3 V algebraically, using chosen loop direction

Algebraic addition of voltages for batteries in series comes directly from Kirchoff’s loop rule.

Why would you want to connect batteries in series?

More voltage! Brighter flashlights, etc. Chemical reactions in batteries yield a fixed voltage. Without changing the chemical reaction (i.e., inventing a new battery type), the only way to change voltage is to connect batteries in series.

This applies to any source of emf, not just batteries!

Go to www.howstuffworks.com to see how batteries work.They even expose the secret of the 9 volt battery!

Click on the picture above only if you are mature enough to handle this graphic exposé.

Go to www.howstuffworks.com to see how batteries work.They even expose the secret of the 9 volt battery!

Shocking!

Six 1.5 V batteries in series!

Why would you want to connect batteries in series the “wrong” way?

You probably don’t want to.

Use could use one battery to charge another—doesn’t seem too useful to me, although might be in special cases.But remember, Kirchoff’s loop rule applies to all emf’s.

You could connect a source of emf – like the alternator in your car – so that it charges a battery.

Rechargeable batteries use an ac to dc converter as a source of emf for recharging.

Could you connect batteries (or sources of emf) in parallel?Sure!

3 V

3 V

You would still have a 3 V voltage drop across your resistor, but the two batteries in parallel would “last” longer than a single battery.

You could use Kirchoff’s rules to analyze this circuit and show that Vab = 3 V.

a b

19.6 Circuits Containing Capacitors in Series and in Parallel

Capacitor:C

Capacitors connected in parallel:

C1

C2

C2

+ -

V

The voltage drop from a to b must equal V.

a b

Vab = V = voltage drop across each individual capacitor.

Vab

C1

C2

C3

+ -

V

a

OSE: Q = C V

Q1 = C1 V

& Q2 = C2 V

& Q3 = C3 V

Now imagine replacing the parallel combination of capacitors by a single equivalent capacitor.

By “equivalent,” we mean “stores the same total charge if the voltage is the same.”

Ceq

+ -

V

a

Q1 + Q2 + Q3 = Ceq V = Q

Q3

Q2

Q1

+ -

Q

Q1 = C1 V Q2 = C2 V Q3 = C3 V

Q1 + Q2 + Q3 = Ceq V

Summarizing the equations on the last slide:

Using Q1 = C1V, etc., in the second line gives

C1V + C2V + C3V = Ceq V

C1 + C2 + C3 = Ceq (after dividing both sides by V)

Generalizing:

OSE: Ceq = Ci (capacitors in parallel)

Does this remind you of any of our resistor equations?

See Giancoli’s comment on why this makes sense, p. 569.

C1

C2

C2

+ -

V

a b

Capacitors connected in series:

C1 C2

+ -

V

C3

An amount of charge +Q flows from the battery to the left plate of C1. (Of course, the charge doesn’t all flow at once).

+Q -Q

An amount of charge -Q flows from the battery to the right plate of C3. Note that +Q and –Q must be the same in magnitude but of opposite sign.

C1 C2

+ -

V

C3

+QA

-QB

The charges +Q and –Q attract equal and opposite charges to the other plates of their respective capacitors:

-Q +Q

These equal and opposite charges came from the originally neutral circuit regions A and B.

Because region A must be neutral, there must be a charge +Q on the left plate of C2.

Because region B must be neutral, there must be a charge --Q on the right plate of C2.

+Q -Q

C1 C2

+ -

V

C3

+QA

-QB

Here’s the circuit after the charges have moved and a steady state condition has been reached:

-Q +Q+Q -Q

Q = C1 V1 Q = C2 V2 Q = C3 V3

The charges on C1, C2, and C3 are the same, and are

But we don’t know V1, V2, and V3 yet.

a b

We do know that Vab = V and also Vab = V1 + V2 + V3.

V3V2V1

Ceq

+ -

V

+Q -QV

Let’s replace the three capacitors by a single equivalent capacitor.

By “equivalent” we mean V is the same as the total voltage drop across the three capacitors, and the amount of charge Q that flowed out of the battery is the same as when there were three capacitors.

Q = Ceq V

Collecting equations:

Q = C1 V1 Q = C2 V2 Q = C3 V3

Vab = V = V1 + V2 + V3.

Q = Ceq V

Substituting for V1, V2, and V3:1 2 3

Q Q QV = + +

C C C

Substituting for V:eq 1 2 3

Q Q Q Q = + +

C C C C

Dividing both sides by Q:eq 1 2 3

1 1 1 1 = + +

C C C C

Generalizing:

OSE: (capacitors in series)

Does this remind you of any of our resistor equations?

ieq i

1 1 =

C C

Example 19-9

Determine the capacitance of a single capacitor that will have the same effect as the combination shown. Use C1 = C2 = C3 = C.

C3

C2

C1

I don’t see a series combination of capacitors, but I do see a parallel combination.

C23 = C2 + C3 = C + C = 2C

C1= CC23 = 2C

Now I see a series combination.

eq 1 23

1 1 1 = +

C C C

eq

1 1 1 2 1 3 = + = + =

C C 2C 2C 2C 2C

eq

2C = C

3

19.7 Circuits Containing A Resistor and a Capacitor

We won’t analyze complex circuits involving many emf’s, resistors, and capacitors in series and parallel.

We will look at simple circuits involving a resistor and a capacitor together. Such a circuit is called an RC circuit. (The circuit to the right is the same as the one in Fig. 19-17.)

+-

R C

S

+-

R C

S

When switch S is closed, current flows.

+Q-Q

However, the flow of current is not instantaneous, but takes time.

The voltage across the capacitor will eventually equal the battery voltage.

As a result, the voltage across the capacitor increases with time.

- t / RCOSE: V = 1 - eεThis equation can easily be derived using a bit of calculus. Ask if you want to see the derivation.

- t / RCV = 1 - eε

The plot to the right shows the voltage vs. time for a 1 F capacitor in series with a 10 resistor and a 10 V battery.Will the voltage across the capacitor “ever” reach 10 V?Note the RC in the exponent in the equation. RC has units of time and is called the “time constant” of the circuit. When t = RC, V = (1 – 1/e) = 0.63 . The voltage across the capacitor reaches 63% of the battery voltage within a time RC.In the circuit for the graph, RC = (10)(10-6) = 10-5 = 10 s.

t = RC

The plot to the right shows the voltage vs. time for C = 1 F and = 10V. For the blue curve, R = 10 and for the “violet” curve, R = 1.Notice how much more rapidly the capacitor charges in the circuit with smaller RC.

Why did we jump to an RC circuit and not study the time behavior of a circuit with only a capacitor?

All circuits have some resistance, if only in the wires and the emf source. Therefore “all” circuits with capacitors are RC circuits.

If the battery is removed from our circuit, the capacitor discharges through the resistor.

R C

S

+Q-Q

You can show that the voltage across the capacitor decays exponentially:

- t / RC0OSE: V = V e

The plot to the right shows the voltage vs. time for our circuit.

R = 10 ; RC = 10 s

R = 1 ; RC = 1 s

Example 19-10

If a charged capacitor, C = 35 F, is connected to a resistance, R = 120 , how much time will elapse until the voltage falls to 10 percent of its original (maximum) value?

- t / RC0OSE: V = V e

1- t / RC1 0V = V e

1- t / RC1

0

V = e

V

1- t / RC1

0

Vln = ln e

V

After a time t1 the voltage drops to a value V1:

t1 is what we want to solve for

1- t / RC1

0

Vln = ln e

V

copied from previous page

11

0

Vln = - t / RC

V

11

0

V t = -RC ln

V

-6 01

0

0.1 V t = - 120 35×10 ln

V

-31 t = 9.7×10 s

19.8 Heart Pacemakers

19.9 Electric Hazards: Leakage Current

Please read this section!

If you value your life, you will read this section!

+ -

V

R

C

Handy stuff to cut and paste: