Ch. 28
wire
open switch
closed switch
2-way switch
1.5 V
+
–
47 F
4.7 k
These circuit elements and many
others can be combined to
produce a limitless variety of
useful devices
•Two devices are in series if they
are connected at one end, and
nothing else is connected there
ideal battery
capacitor
resistor
•Two devices are in parallel if
they are connected at both ends
Resistors in Parallel and in Series
•When resistors are in series, the same
current must go through both of them
•The total voltage difference is
R1
R2
V1  IR1
V2  IR2
V  V1  V2  I  R1  R2 
R  R1  R2
•The two resistors act like one with resistance
•When resistors are in parallel, the same potential
is across both of them
•The total current through them is
V V

I  I1  I 2 
R1
R2
•The two resistors act like one with resistance
V
R
I
 1
1 



 R1 R2 
1
R1
R2
V  I1 R1  I 2 R2
1
1
1


R R1 R2
Warmup 10x
q1
Parallel and Series - Formulas
Capacitor
Series
Resistor
Inductor*
1
1
1


R  R1  R2
C C1 C2
L  L1  L2
1
1
1


R R1 R2
1
1
1


L L1 L2
V  IR
dI
EL   L
dt
Parallel
C  C1  C2
Fundamental
Formula
Q
V 
C
* To be defined in a later chapter
The Voltage Divider
•Many circuits can be thought of as a voltage divider
•Intentionally or unintentionally
What’s the voltage drop across each of the resistors?
R1
+
–
E
R2
R  R1  R2
R1
V1  IR1 
E
E
R1  R2
I
R1  R2
R2
V2  IR2 
E
The larger resistor gets
R1  R2
most of the voltage
If Mr. Curious has a resistance of 10 k
and the light bulb has a resistance of 240
, how bright is Mr. Curious?
120 V
+
–
Vcurious
10000

E=
 117 V
10240
Not very bright
CT – 1 Consider two identical resistors wired in series (one behind the other). If there is
an electric current through the combination, the current in the second resistor is
A. equal to
B. half
C. smaller than, but not necessarily half
the current through the first resistor.
Ans A
CT – 2 As more identical resistors R are added to the parallel circuit shown here, the
total re-sistance between points P and QB
Ans C
A. increases.
B. remains the same.
C. decreases.
CT – 3 Charge flows through a light bulb. Suppose a wire is connected across the bulb as
shown. When the wire is connected,
A.
B.
C.
D.
Practically all the charge continues to flow through the bulb.
half the charge flows through the wire, the other half continues through the bulb.
Practically all the charge flows through the wire.
none of the above
Ans C
CT - 4-The circuit below consists of two identical light bulbs burning with equal
brightness and a single 12 V battery. When the switch is closed, the brightness of bulb A
A. increases.
B. remains unchanged.
C. decreases.
Ans A
CT - 5-If the four light bulbs in the figure are identical, which circuit puts out more light?
A. I.
B. The two emit the same amount of light.
C. II.
Ans A
Ex- (a) Find the equivalent resistance between points a and b in the figure below. (b) If a
potential difference of 34 V is applied between points a and b, calculate the current in
each resistor.
Solve on Board
Ideal vs. Non-Ideal Batteries
+
•Up until now, we’ve treated a battery
E
as if it produced a fixed voltage, no
matter what we demand of it
ideal battery
•Real batteries also have resistance
r
E
•It limits the current and therefore
realistic battery
the power that can be delivered
•If the internal resistance r is small
The maximum potential
compared to other resistances in the
difference E across the
problem, we can ignore it
battery is called
10  30 V
electromotive force (emf)
+
–
–
+
–
A 30 V battery with 10  of resistance is
50 
connected to a 50  resistor. What is the
actual voltage across the 50  resistor?

r 
B) 36 V
C) 6 V
V   1 
 E  25 V A) 30 V
rR 

D) 25 V
E) 24 V
JIT
Ans: (i) b (ii) a
Ans: (iii) a (ii) b
Kirchoff’s First Law
The total current into any vertex equals the current out of that vertex
Which equation do you get for point A?
A) I1 + I2 = I3
B) I2 + I3 = I1
C) I1 + I3 = I2
D) I1 + I2 + I3 = 0
•The equation from point B is
I 3  I1  I 2
5
4
You always get one
redundant equation
– A
6V
+
I2
B
I1
3
 Iin   Iout
12 V
–
+
How to apply it:
•First, assign a current and a direction to every pathway
•Two components in series will always have the
same current
•At every vertex, write the equation:
I3
Kirchoff’s Second Law
The total voltage change around a loop is always zero
0  18  3I1  5I 2
I1
3
0  12 3I1 6 5I 2
12 V
–
+
I2
+
How to apply it:
•First, assign a direction to every loop
•I often pick clockwise
•Start anywhere, and set 0 equal to sum of
potential change from each piece:
•For batteries: V = E
•It is an increase if you go from – to +
•It is a decrease if you go from + to –
•For resistors: V = IR
•It is a decrease if you go with the current
•It is an increase if you go against the current
–
5
6V
4
I3
Kirchoff’s Second Law (2)
I 3  I1  I 2
Three equations in three unknowns: solve it
•We can let Maple do it for us
> solve({i3=i1+i2,0=-5*i2-6.-4*i3,
0=18-3*i1+5*i2},[i1,i2,i3]);
Negative currents means we guessed the wrong way
•Not a problem
I2
+
0  18  3I1  5I 2
I1
3
0  5I 2 6 4I 3
12 V
–
+
What is Kirchoff’s Second Law for the purple loop?
A) 0 = +5I2 – 6 – 4I3 B) 0 = +5I2 + 6 – 4I3
C) 0 = –5I2 – 6 – 4I3 D) 0 = –5I2 + 6 – 4I3
–
5
6V
4
I3
Ex- Using Kirchoffs' rules, (a) find the current in each resistor shown below. (b) Find the
potential difference between points c and f. Which point is at a higher potential?
Solve on Board
Kirchoff’s Laws with CapacitorsI
If know which side is positive then that is high potential – work
like battery.
Q
C
•The voltage change is given by V = Q/C
•It is a decrease if (+)Q is the side you are going in +
•It is an increase if Q is the side you are going out –
•The current is related to the time change of Q
• Add minus sign if I doesn’t enter from the same side
dQ
I
as Q – it is minus if decreasing
dt
•If you are in a steady state, the current through a
capacitor is always zero
In this circuit, in the
steady state, where is
current flowing?
+
–
+
–
It’s really just a battery
and two resistors in series!
The Simplest RC Circuit
Q0
R
I
C
In the circuit shown at left, the capacitor starts
with charge Q0. At time t = 0, the switch is
closed. What happens to the charge Q?
•Current begins to flow around the loop, so the charge Q will change
Q
0   RI
C
dQ
Q
 I  
dt
RC
•This is a differential equation, and therefore hard to solve
dQ
dt

Q
RC
Q  et RC
Q  Q0 e
 t RC
dQ
dt
 Q   RC
Check the units:
RC
 F
t
ln Q  
k
RC
V C

A V
C
C s
s
Charging and Discharging Capacitors
•The combination RC =  is called the time constant
•It’s the characteristic time it takes to discharge
•We can work out the current from
C
RC
Q  Q0 e t 
  RC
dQ Q0
t 

e
I 

dt

C

t 
0

e
 0 e t 
RC
R
Q
Q  Q0 e t
I
R
dQ
I
dt
Q
0   IR  E
C
+
–
E
In this circuit, the capacitor is initially
uncharged, but at t = 0 the switch is closed
dQ
Q
E


dt
RC R
Q  EC 1  e t
RC

Warmup 11
CT – 6 A simple circuit consists of a resistor R, a capacitor C charged to a potential Vo
(not shown), and a switch that is initially open but then thrown closed. Immediately after
the switch is thrown closed, the current in the circuit is
A. V o /R.
B. zero.
C. need more information
Ans A
JIT Quick Quiz 28.5
Consider the circuit in the figure and assume that the battery has no internal
resistance.
(i)
Just after the switch is closed what
is the current in the battery (a) 0
(b) /2R,(c) 2/R, (d) /R, (e)
impossible to determine
(ii) After a long time, what is the
current in the battery?
Ans i) c, ii) d
Example (Serway 28-38). Consider a RC circuit consisting of a Emf = 30.0V, a resistor
= 1.00 Ma capacitor = 5.00 F and a switch like in Figure 28.34 (charging a
capacitor). Find (a) the time constant of the circuit and (b) the maximum charge on the
capacito after the switch is closed. (c) If the switch is closed at t = 0, find the current in
the resistor 10.0 s later.
Solve on Board
Ammeters and Voltmeters
•An ammeter is a device that measures the current (amps) anywhere
in a circuit
A
•To use it, you must route the current through it
•A perfect ammeter should have zero resistance
•A voltmeter is a device that measures the potential difference (volts)
between any two points in a circuit
V
•To use it, you can simply connect to any two points
•A perfect voltmeter has infinite resistance
A
Which meter is installed incorrectly?
A) Left voltmeter
B) Right voltmeter
+
V
A C) Left ammeter
D) Right ammeter
–
E) All are correct
V
•Voltmeters should be connected to two places in an existing circuit
•The left voltmeter is placed correctly
•A voltmeter has infinite resistance
•The right one effectively blocks the current on the right
Household Wiring
*Actually, this
is alternating
current, later
chapter
•All household appliances consume electrical power
•Think of them as resistors with fixed resistance R
•Devices are designed to operate at 120 V*
2
•Often, they give the wattage at this voltage
P   V  R
•Can easily get the effective resistance from
•To make sure power is given to each device, they are all placed in
parallel
Fuse
A
+
box
–
Inside House
•If you put too many things on at once, a lot of current is drawn
•The wires, which have some resistance, will start to get hot
•To avoid setting the house on fire, add a fuse (or a circuit breaker)
Warmup 11
Why three wires?
•If a device is functioning properly, you need only two wires
•“Live” and “Neutral” wires
Toaster
•If the live wire accidentally touches the casing, the person
can be electrocuted
•The wrong solution – connect the neutral to the casing
•Now imagine the neutral wire breaks
•The person again can be electrocuted
•The right solution: Add a third “ground” wire connected
directly to ground
•Normally no current will flow in this wire
•If the hot wire touches the casing, it will trigger the
fuse/circuit breaker and protect the person