p. 39
Resistors Connected in Series and in Parallel
Consider the circuit shown in the figure at the right. The two resistors
with resistances R1 and R2 are connected together and then the two
free ends (one from the first resistor and the other from the second
resistor) are connected across a seat of emf. Resistors wired like this
are said to be connected in series.
I
The two resistors connected together can be thought of as a single
equivalent resistance R, as shown in the second figure below. We
would like to find the relationship between R and the individual
resistances R1 and R2.
I
V1
V2
Assume that the seat of emf is
ideal (that is, has no internal
resistance). Let the current
drawn from the seat of emf be I. Then the voltage drops across the
individual resistors are
V1  IR1 and V2  IR2 .
The sum of these voltage drops must be the voltage drop across the seat
of emf:
V  V1  V2 or V  IR1  IR2 .
However, interpreting these two resistors connected in series as equivalent to a single resistance R we must also
have
V  IR
so that
IR  IR1  IR2 , from which
R  R1  R2
This is the formula for the equivalent resistance of two resistors connected in series.
p. 40
Resistors Connected in Series and in Parallel (continued)
In the last lecture we showed that if two resistors R1 and R2 are connected in series the equivalent resistance is
given by
R = R1 + R2.
This formula is easily extended to the case where three, four or more resistors are connected in series.
A set of resistors are connected in series if and only if the current through each resistor is the same. This is
possible only if one resistor follows another in a connection sequence.
Resistors R1 and R2 are connected in series.
Resistors R1 and R2 are not connected in series. (Why?)
Are resistors R2 and R3 connected in series? How can you tell?
Example
Resistor R1 = 12.0 ohms and R2 = 8.00 ohms are connected in series across an ideal dry cell battery. (An ideal
battery has no internal resistance.) A voltmeter measures a 7.50 volt potential drop across R1.
a. What is the current in the circuit?
V1  IR1; I 
V1
7.50
; I
R1
12.0
I  0.625 A
b. What is the voltage of the battery?
The equivalent resistance of the resistors is R = 20 ohms.
The current I through R1 is also the current through the battery: V  IR; V   0.625 A  20 ohms 
V = 12.5 volts
p. 41
c. (Alternative solution to b.) The potential drop (or voltage or voltage drop) across R2 is
V2  IR2 ; V2   0.625 A  8.00 ohms  ; V2  5.00 volts
The voltage across the battery is the sum of the voltage drops across the resistors:
V  V1  V2 ; V  7.5 volts + 5.00 volts; V  12.5 volts
(The same answer as before.)
Parallel Wiring
The figure at the right shows resistors R1 and R2 connected in parallel.
A set of resistors is connected in parallel if and only if the voltage across each
resistor is the same.
If a battery with voltage V is placed across resistors connected in
parallel, the battery “sees” a single equivalent resistance R placed
across its terminals. To find this equivalent resistance we need
Kirchhoff’s junction rule:
I
V
I1
I2
The total current entering a junction equals the total current
leaving that junction.
Thus I  I1  I 2 and
V V V
1 1
1
 
. The V's cancel so that   . This formula is easily extended to the case of three or
R R1 R2
R R1 R2
more resistors connected in parallel.
Example
The 12.0 ohm and 8.00 ohm resistors of the previous example are connected in parallel across the same ideal
dry cell battery.
a. What is the current drawn from the battery?
First find the equivalent resistance R.
1
1
1


R 12.0  8.00 
 8.33  102  1  12.5  10 2  1
 20.8 102  1
R  4.81 
I
V
12.5 V
; I
; I  2.60 A
R
4.81 
p. 42
b. Find the current through each resistor.
V
12.5 V
I1  ; I1 
; I1  1.04 A
R1
12.0 
I2 
V
12.5 V
; I2 
; I 2  1.56 A
R2
8.00 
Check : I1  I 2  2.60 A
Equivalent Resistance of More Complicated Networks
How are resistances R2 and R3 connected in the resistance network
shown at the right? They are not connected in series, because the
current through R2 is not necessarily the current through R3 (why?)
R2 and R3 are connected in parallel because the potential drop across
R3 is the same as the potential drop across R2. (Note that we can go
from the top node to the bottom node in two ways: the first way is to
just cross R3; the second is to just cross R2. Hence the potential drop
across R3 must be the same as that across R2.)
a

b
Let R1 = R1 = 2.0 , R2 = R2 = 2.0 , R3 = R3 = 6.0 .
First find the equivalent resistance of R2 and R3:
1
1
1
1
1
1

 ;


;
R23 R2 R3 R23 3.0  6.0 
1
3

R23 6.0 
R23  2.0 
Now R23 is in series with R1 , so that the equivalent resistance is
R  R1  R23 ;
R  2.0   2.0 ; R  4.0 
a
b
Here is an example from another textbook. Can you do this problem?
Here are the answers: (a) 8.0 ohms, (b) 12 volts.
p. 43
Internal Resistance
All real seats of emf have an internal resistance. If no current is drawn from the seat of emf, the internal
resistance has no effect. When the seat of emf is connected to a circuit and current is drawn from it, the
potential difference across its terminals is less than when no current is drawn from it. The actual voltage
between the terminals of a seat of emf is called the terminal voltage. A hypothetical seat of emf with zero
internal resistance is called an ideal seat of emf.
r
E
A real seat of emf can be thought of as an ideal seat of emf connected in
series with an internal resistance r. E is the emf when no current is drawn.
When a current I is drawn there is a potential drop Ir across the internal
resistance. The terminal voltage of the seat of emf is then
V  E  Ir
Example
A 3.00 ohm resistor is connected to a 12.0 volt battery with an internal resistance of 0.200 ohms.
a. How much current is drawn from the battery?
E
The equivalent resistance of the circuit is
R = 3.00  + 0.200 ; R = 3.20 .
r
3.0 
The current drawn from the seat of emf is then
I
E
12.0 V
; I
; I  3.75 A
R
3.20 
b. What is the terminal voltage of the battery?
V  E  Ir
 12.0 V   3.75 A  0.200  
V  11.3 V
In general, the more current that is drawn from a seat of emf the lower its terminal voltage due to its internal
resistance.