Chapter 19
DC Circuits
Units of Chapter 19
• EMF and Terminal Voltage
• Resistors in Series and in Parallel
• Kirchhoff’s Rules
• EMFs in Series and in Parallel; Charging a
Battery
• Circuits Containing Capacitors in Series
and in Parallel
Units of Chapter 19
• RC Circuits – Resistor and Capacitor in
Series
• Electric Hazards
• Ammeters and Voltmeters
19.1 EMF and Terminal Voltage
Electric circuit needs battery or generator to
produce current – these are called sources of
emf.
Battery is a nearly constant voltage source, but
does have a small internal resistance, which
reduces the actual voltage from the ideal emf:
(19-1)
19.1 EMF and Terminal Voltage
This resistance behaves as though it were in
series with the emf.
Example 19-1
A 65.0 Ω resistor is connected to the terminals of a battery whose emf is
12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in
the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power
dissipated in the battery’s internal resistance r.
(a) Vab =  - Ir and V ab = IR so IR =  - Ir

12.0 V
I=
=
= 0.183 A
R +r 65.0  +0.5 
(b) Vab =  - Ir =12.0 V - (0.183 A)(0.5 ) =11.9 V
(c) PR = I 2R = (0.183 A) 2 (65.0 ) = 2.18 W
Pr = I 2r = (0.183 A) 2 (0.5 ) = 0.02 W

19.2 Resistors in Series and in Parallel
A series connection has a single path from
the battery, through each circuit element in
turn, then back to the battery.
19.2 Resistors in Series and in Parallel
The current through each resistor is the same;
the voltage depends on the resistance. The
sum of the voltage drops across the resistors
equals the battery voltage.
(19-2)
19.2 Resistors in Series and in Parallel
From this we get the equivalent resistance (that
single resistance that gives the same current in
the circuit).
(19-3)
19.2 Resistors in Series and in Parallel
A parallel connection splits the current; the
voltage across each resistor is the same:
19.2 Resistors in Series and in Parallel
The total current is the sum of the currents
across each resistor:
19.2 Resistors in Series and in Parallel
This gives the reciprocal of the equivalent
resistance:
(19-4)
19.2 Resistors in Series and in Parallel
An analogy using
water may be helpful
in visualizing
parallel circuits:
Example 19-3
Two 100 Ω resistors are connected (a) in parallel, and (b) in series, to a
24.0 V battery. What is the current through each resistor and what is the
equivalent resistance of each circuit?
V V 24.0 V 24.0 V
+
=
+
= 0.48 A
R1 R 2 100  100 
V 24.0 V
R eq = =
= 50 
I 0.48 A
(b) V = V1 + V2 = IR1 + IR 2
V
24.0 V
I=
=
= 0.120 A
R1 + R 2 100  +100 
V 24.0 V
R= =
= 200 
I 0.120 A
(a) I = I1 + I 2 =

Example 19-4
How much current is drawn from the battery shown in the figure?
1
1
1
=
+
= 0.0034 1
R P 500  700 
1
= 290 
1
0.0034 
R eq = 400  +290  = 690 
RP =
I=

V 12.0 V
=
= 0.017 A =17 mA
R eq 690 
Example 19-5
What is the current through the 500 Ω resistor in the figure?
Vab = IR = (0.017 A)(400 ) = 7.0 V
Vbc = 12.0 V - 7.0 V = 5.0 V
5.0 V
I1 =
= 1.0x10 -2 A = 10 mA
500 
5.0 V
I2 =
= 7.0x10 -3 A = 7 mA
700 

Example 19-7
A 9.0 V battery whose internal resistance r is 0.50 Ω is connected in the
circuit shown. (a) How much current is drawn from the battery? (b) What is
the terminal voltage of the battery? (c) What is the current in 6.0 Ω resistor?
(a)
3
1
1
1
so R = 2.7 
=
+
=
R eq1 8.0  4.0  8.0 
R eq2 = 6.0  + 2.7  = 8.7 
1
1
1
= 0.21 1 so R = 4.8 
+
=
R eq3 10.0  8.7 
R eq = 4.8  + 5.0  + 0.50  = 10.3 
I=

R eq
=
9.0 V
= 0.87 A
10.3 
(b) Vab =  - Ir = 9.0 V - (0.87 A)(0.50 ) = 8.6 V
(c) Current across 6.0  resistor is same as that
across 8.7  resistor.
V8.7 = 9.0 V - (0.87 A)(0.50  + 5.0 )
9.0 V - (0.87 A)(0.50  + 5.0 )
= 0.48 A
I' =
8.7 
19.3 Kirchhoff’s Rules
Some circuits cannot be broken down into
series and parallel connections.
19.3 Kirchhoff’s Rules
For these circuits we use Kirchhoff’s rules.
Junction rule: The sum of currents entering a
junction equals the sum of the currents
leaving it.
19.3 Kirchhoff’s Rules
Loop rule: The sum of
the changes in
potential around a
closed loop is zero.
19.3 Kirchhoff’s Rules
Problem Solving: Kirchhoff’s Rules
1. Label each current.
2. Identify unknowns.
3. Apply junction and loop rules; you will
need as many independent equations as
there are unknowns.
4. Solve the equations, being careful with
signs.
Example 19-8
Calculate the currents I1, I2, and I3 in the three branches of the circuit shown .
Junction rule : I 3 = I1 + I 2
Loop rule : - 30I1 + 45 - (40 +1)I 3 = 0
Loop rule : - 30I1 + (20 +1)I 2 - 80 = 0
From the 3rd equation, we have
80 + 30I1
= 3.8 +1.4I1
21
From the 2nd equation, we have
I2 =
45 - 30I1
= 1.1- 0.73I1
41
Using these in the first equation,
I3 =
I1 = I 3 - I 2 = 1.1- 0.73I1 - 3.8 -1.4I1
I1 = -0.87 A
I 2 = 3.8 +1.4I1 = 3.8 +1.4(-0.87) = 2.6 A
I 3 = 1.1- 0.73I1 = 1.1- 0.73(-0.87) = 1.7 A
19.4 EMFs in Series and in Parallel;
Charging a Battery
EMFs in series in the same direction: total
voltage is the sum of the separate voltages
19.4 EMFs in Series and in Parallel;
Charging a Battery
EMFs in series, opposite direction: total
voltage is the difference, but the lowervoltage battery is charged.
19.4 EMFs in Series and in Parallel;
Charging a Battery
EMFs in parallel only make sense if the
voltages are the same; this arrangement can
produce more current than a single emf.
Example 19-9
A good car battery is being used to jump start a car with a weak battery. The good
battery has an emf of 12.5 V and internal resistance 0.020 Ω. Suppose the weak battery
has an emf of 10.1 V and internal resistance 0.10 Ω. Each copper jumper cable is 3.0 m
long and 0.50 cm in diameter, and can be attached as shown. Assume the starter motor
can be represented as a resistor Rs=0.15 Ω. Determine the current through the starter
motor (a) if only the weak battery is connected to it, and (b) if the good battery is also
connected.
(a) R eq = 0.10  + 0.15  = 0.25 
I = V/R = (10.1 V)/(0.25 ) = 40 A
L
(3.0 m)
(b) R J =  = (1.68x10 -8 m)
= 0.0026 
A
(0.25x10 -2 m) 2
Loop rule : 12.5 V - I1 (2R J + r1) - I 3R s = 0
12.5 V - I1 (0.025 ) - I 3 (0.15 ) = 0
Loop rule : 10.1 V - I 3 (0.15 ) - I 2 (0.10 ) = 0
Junction rule : I1 + I 2 = I 3
I 3 = 71 A, I 2 = -5A, I1 = 76 A
19.5 Circuits Containing Capacitors in
Series and in Parallel
Capacitors in
parallel have the
same voltage across
each one:
19.5 Circuits Containing Capacitors in
Series and in Parallel
In this case, the total capacitance is the sum:
(19-5)
19.5 Circuits Containing Capacitors in
Series and in Parallel
Capacitors in series have the same charge:
19.5 Circuits Containing Capacitors in
Series and in Parallel
In this case, the reciprocals of the
capacitances add to give the reciprocal of the
equivalent capacitance:
(19-6)
Example 19-10
Determine the capacitance of a single capacitor that will have the same effect
as the combination shown in the figure. Take C1=C2=C3=C.
C 23 = C 2 + C 3 = C + C = 2C
1
1
1
1 1
3
=
+
= +
=
C eq C1 C 23 C 2C 2C
2
C eq = C
3
Example 19-11
Determine the charge on each capacitor in the figure, and the voltage across
each, assuming C=3.0 µF and the battery voltage is V=4.0 V.
Q = CV = (2.0 F)(4.0 V) = 8.0 C
This Q arrives at the negative plate of C 1 so C1 = 8.0 C.
The charge that leaves the positive plate is split between
C 2 and C 3 , so they each equal 0.5Q 1 = 4.0 C.
V1 = Q1/C1 = (8.0 C)/(3.0 F) = 2.7 V
V2 = Q 2 /C 2 = (4.0 C)/(3.0 F) = 1.3 V
V3 = Q 3 /C 3 = (4.0 C)/(3.0 F) = 1.3 V
19.6 RC Circuits – Resistor and Capacitor
in Series
When the switch is closed, the capacitor will
begin to charge.
19.6 RC Circuits – Resistor and Capacitor
in Series
The voltage across the capacitor increases
with time:
This is a type of exponential.
19.6 RC Circuits – Resistor and Capacitor
in Series
The charge follows a similar curve:
This curve has a characteristic time constant:
(19-7)
19.6 RC Circuits – Resistor and Capacitor
in Series
If an isolated charged capacitor is connected
across a resistor, it discharges:
Example 19-12
If a charged capacitor, C= 35 µF, is connected to a resistance R=120 Ω as in
the figure, how much time will elapse until the voltage falls to 10% of its original
(maximum) value?
 = RC = (120 )(35x10 -6 F) = 4.2x10 -3 s
VC = V0 (e -t/RC) so 0.10V0 = V0e -t/RC
-t
e
= 0.10 and ln(e
)=
= ln0.10 = -2.3
RC
t = 2.3(RC) = (2.3)(4.2x10 -3 s) = 9.7x10 3 s = 9.7 ms
-t/RC
-t/RC
19.7 Electric Hazards
Even very small currents – 10 to 100 mA can
be dangerous, disrupting the nervous system.
Larger currents may also cause burns.
Household voltage can be lethal if you are wet
and in good contact with the ground. Be
careful!
19.7 Electric Hazards
A person receiving a
shock has become part
of a complete circuit.
19.7 Electric Hazards
Faulty wiring and improper grounding can be
hazardous. Make sure electrical work is done by
a professional.
19.7 Electric Hazards
The safest plugs are those with three prongs;
they have a separate ground line.
Here is an example of household wiring – colors
can vary, though! Be sure you know which is the
hot wire before you do anything.
19.8 Ammeters and Voltmeters
An ammeter measures current; a voltmeter
measures voltage. Both are based on
galvanometers, unless they are digital.
The current in a circuit passes through the
ammeter; the ammeter should have low
resistance so as not to affect the current.
19.8 Ammeters and Voltmeters
A voltmeter should not affect the voltage across
the circuit element it is measuring; therefore its
resistance should be very large.
19.8 Ammeters and Voltmeters
An ohmmeter measures
resistance; it requires a
battery to provide a
current
19.8 Ammeters and Voltmeters
If the meter has too
much or (in this case)
too little resistance, it
can affect the
measurement.
Summary of Chapter 19
• A source of emf transforms energy from
some other form to electrical energy
• A battery is a source of emf in parallel with an
internal resistance
• Resistors in series:
Summary of Chapter 19
• Resistors in parallel:
• Kirchhoff’s rules:
1. sum of currents entering a junction
equals sum of currents leaving it
2. total potential difference around closed
loop is zero
Summary of Chapter 19
• Capacitors in parallel:
• Capacitors in series:
Summary of Chapter 19
• RC circuit has a characteristic time constant:
• To avoid shocks, don’t allow your body to
become part of a complete circuit
• Ammeter: measures current
• Voltmeter: measures voltage
Homework - Ch. 19
• Questions #’s 1, 2, 4, 5, 15, 16
• Problems #’s 3, 9, 11, 13, 17, 19, 21, 23,
25, 27, 29, 31, 35, 37, 41, 43, 51