# Document ```Fundamentals of DC Circuits
(Chapter 32)
What types of circuits will we study here?
DC CIRCUITS (DIRECT CURRENT CIRCUITS)!
• DC circuits have a source of emf (eg. battery) that
does not change with time
• The current can vary with time but the direction of the
current does not change.
In Chapter 36 we will learn AC (Alternating Current)
Circuits. Alternating currents are created by electrical
generators.
A circuit diagram is a short hand pictorial representation of the
components in a circuit and how they are connected.
(In Chapter 36, we will add symbols for inductors and generators)
Examples of Circuit Diagram
=
=
4 Rules for Solving DC Circuits:
1)Kirchoff's Junction Law:
!I
in
= ! I out
2)Kirchoff's Loop Law:
&quot;Vclosed #loop = ! &quot;Vi = 0
i
3)Ohm's Law (applies to resistors and light bulbs)
&quot;V = IR = R(dQ / dt)
4)The voltage across a capacitor
&quot;VC = Q / C
Sign Conventions for Kirchoff s Loop Law
Batteries:
!V = Va &quot; Vb
(going uphill )
(going downhill )
Sign Convention for Resistors
Moving across the resistor in the direction of current decreases
!V = Va &quot; Vb
the potential
electron s rolling
downhill
Sign Convention for Capacitors
( uphill like battery)
( downhill )
Analysis of a Simple Circuit
1.  Current is the same everywhere
(There are no junctions!)
2. Assume the wires are ideal .
This means we can ignore their
resistance.
Solution
0 = !Vloop = !Vbattery + !VR1 + !VR2
0 = &quot; + (#IR1 ) + (#IR2 )
I = &quot; / (R1 + R2 )
I = 0.50A
!VR1 = #IR1 = #5.0V
!VR2 = #IR2 = #10.0V
Simulations....
Energy and Power in Electric Circuits
A circuit is essentially a device that transforms energy into different
forms starting from chemical energy in a battery.
Chemical Energy of Battery
Electrical Potential Energy
Kinetic Energy of Electrons
Thermal Energy
(Light Bulbs Get Hot
and Glow!)
Power Delivered by Battery (EMF Source)
The battery raises the potential energy of electrons.
Power= rate at which battery transfers energy to electrons
!U = Q!Vbattery
P=
dE
(General defn. of power as time rate of change of energy)
dt
d
&quot; dQ %
Q!Vbattery = \$
!Vbattery
'
# dt &amp;
dt
= I !Vbattery (Power supplied by battery to circuit)
Pbattery =
Pbattery
This is the rate at which chemical energy is transformed into
electrical potential energy!
Power Dissipated Due to Heating
The electric potential energy causes electrons to move downhill .
Potential Energy --&gt;Kinetic Energy
Kinetic Energy is transformed into heat via collisions (resistance=electron friction )
Kinetic energy obtained from E field made by battery:
!K = W = F!x = (qE)d (d=distance between collisions)
Kinetic energy transfered to lattice when electron
collides with atom:
!E collision = !K = qEd
Over a distance L, the electron collides N=L/d times
!E Thermal = N !Ecollision = qE(Nd) = q!V
Power dissipated by resistor/lightbulb:
dE
dq
PR = Thermal =
!VR = I !VR
dt
dt
Energy and Power
Power is the energy supplied or expended per unit time
P=dE/dt. Unit of power is the Watt 1W=1J/s.
The power supplied by a battery to a circuit:
Psupplied = I!Vbattery
The power expended by an Ohmic device (resistor, wire,
light bulb):
Pexpended = I!VR = I
2
!VR )
(
R=
R
2
The power dissipated by two series resistors
How much power do each of the resistors dissipate ?
!=
Solution
Kirchoff's Law for the Voltage:
! &quot; IR1 &quot; IR2 = 0
!
I=
= 0.40A
(R1 + R2 )
2
2
W
PR1 = I R1 = 0.4A # 12\$ = 1.92W
PR2 = I 2 R2 = 0.4A 2 # 18\$ = 2.9W
Pbattery = 12V ! 0.4A = 4.8W
Multiple Resistors: Series vs. Parallel
Connections
Series: Resistors connected
sequentialy. Single closed current loop.
Parallel: Each
resistor is individually
connected to battery
Resistors in Series:
How to Derive the Equivalent Resistance
&quot; !V = 0
i
i
0 = Vbattery # IR1 # IR2 # IR3
I(R1 + R2 + R3 ) = Vbattery
•  Resistors are aligned end to end,
with no junctions between them,
•  The current I is the same through
all resistors placed in series.
IReq = Vbattery
Req = R1 + R2 + R3 (equivalent resistance)
Parallel Resistors:
How to Derive the Equivalent Resistance
I = I1 + I 2 (Kirchoff's Junction Law at junction a OR b)
I1 = !V1 / R1 = !Vbattery / R1
 The potential differences ΔV are the same
across all resistors placed in parallel.
I 2 = !V2 / R2 = !Vbattery / R2
I=
!Vbattery
R1
+
!Vbattery
R2
=
!Vbattery
Req
&quot;
1
1
1
=
+
Req R1 R2
&quot; Req =
R1 R2
R1 + R2
Summary of Parallel vs. Series
An Example of Resistor Network...
Solution...
A Nontrivial Example:
Initial Circuit:
Equivalent to this circuit:
R/2 resistors are in series
Because there are 2 sources
of emf, we can t reduce this to
a single equivalent resistor !
Is the resistor between a-b in
series or parallel ?
Solution: Kirchoff s Rules Always Work !
There are two separate closed loops that
enclose the segment a-b:
Yields a system of three linear equations.
At the junctions a and b:
I1 = I 2 + I a !b
First Loop:
!2&quot; ! I1 R ! I a !b R = 0
Second Loop:
&quot; + I a !b R = 0
! &quot; ! I1 R = 0
Second Loop + Junction Eq.:
&quot; = (I 2 ! I1 )R
I2 = 0
Comment About the Sign of I:
I1 = ! &quot; / R
We chose the direction for I1 incorrectly!
Kirchoff's Law is &quot;Self-Correcting&quot;.
If you choose the direction of the current wrongly,
the solution will be negative.
Real Batteries
Real batteries have resistance! This internal resistance is usually
small compared to the resistance of the circuits.
Typical internal resistances:
1)Akaline batteries (AA, AAA, C, D): 0.15-0.3 Ohms
3)Lithium Ion (cell phone, laptop): 0.3 Ohms
Internal resistance reduces the
voltage delivered by battery to circuit:
!Vb = &quot; # Ir = &quot; #
!Vb =
R
&quot;
R+r
r
&quot;
R+r
Example of Internal Resistance
What is the voltage of the battery and internal resistance??
Case 1: Switch is open
!
I'=
= 1.636A
r + 5.0&quot;
! = I '(r + 5.0&quot;)
Solution
Case 2: Switch is closed
I = I1 + I 2
I1 = 1.565A
I 2 = I # I1
Resistors are in parallel---&gt; They have the same voltage
I1 (5.0&quot;) = 7.825V = (I # I1 )(10&quot;)
15&quot;
I = I1
= 2.348A
10&quot;
Closed Switch- Loop through ammeter
! # Ir # I1 (5.0&quot;) = 0
! = Ir + I1 (5.0&quot;)
Final Solution...
Two equations and two unknowns:
! = I '(r + 5.0&quot;) = (1.636A)(r + 5.0&quot;)
! = Ir + I1 (5.0&quot;) = (2.348A)r + 7.825V
Solve....
! = 9.0V
r = 0.50&quot;
Electrical Measurements: Ammeters &amp;
Voltmeters
Ammeters
Ammeters measure electric currents I in circuits. They must be
placed in series with the device through which the current is to be measured.
An analog ammeter (aka galvanometer) measures current by magnetic forces.
The current in the wire creates a magnetic field that interacts with the permanent
magnets deflecting the needle. Works on same principle as electric motor.
Ammeter has a small internal resistance that is usually negligible.
Voltmeters
1)A voltmeters is a device used to measure the voltage across
a circuit element.
2) It is connected in parallel with the circuit element to be measured.
3)Connecting in parallel yields the same potential difference
across the voltmeter and circuit element.
What is the ideal
resistance of a
voltmeter???
Voltmeters (2)
An ideal voltmeter has infinite resistance.
The equivalent resistance of the voltmeter and resistor is
1
1
1
1
= +
! when Rvoltmeter ! &quot;
Req R Rvoltmeter
R
When Rvoltmeter ! &quot;, the voltmeter doesn't
disturb the circuit by drawing any current.
Analog voltmeters consist of a very very
large resistor in series with a sensitive
ammeter
An Even More Difficult Example :)
All resistors are R and batteries ε.
What does the voltmeter and
=
???
First Simplification....
R
1
1
1
1
=
+
=
Rab 2R 2R R
R
Solution (Cont.)
Solve like earlier problem.
Voltmeter draws no current
and Ammeter is assumed ideal.
Three Equations for
three unknowns (I1, I2,Iab):
1.  Current Junction
2.  Loop 1
3.  Loop 2
Loop 2
Loop 1
I1 I
ab
I2
Resistor-Capacitor (RC) Circuits: Time Varying
Currents.
•  Resistor and Capacitor
are in series.
•  The resistor restricts the
current and prevents
capacitor from
discharging instantly.
•  Use Kirchoff s loop law
to get a differential
equation for the charge
in the circuit
Case 1: Solving for a Discharging Capacitor
!VC &quot; !VR = 0
Q / C &quot; (&quot;IR) = 0
dQ
I=
dt
dQ
= &quot;Q / #
# =RC
dt
Q = Q0 e&quot;t /# (Charge on Capacitor)
I = I 0 e&quot;t /# (Current through resistor)
Case 2: Charging an RC Circuit With a Battery
•  Initially there is no
charge on the capacitor.
•  Closing the switch
connects the RC to the
battery allowing current
to flow and the
capacitor to charge.
Case 2: Solving for the Charging of an RC
Circuit
! &quot; IR &quot; Q / C = 0
dQ
+Q /# = ! / R
dt
Q(t) = Qmax (1 &quot; e&quot;t /# )
Qmax = ! C
Qmax &quot;t /# \$ ! ' &quot;t /#
I(t) =
e =&amp; )e
% R(
#
Example: Energy Dissipated Through Resistor
When the switch is flipped from a to b, how much energy is
dissipated through the resistor after 1.25ms??
Solution......
The energy dissipated by the resistor equals the change in the
energy stored in the capacitor :
Q(t = 0) = ! C = (50V )(20 &quot; 10 #6 F) = 1.0 &quot; 10 #3
1 Q 2 1 (1.0 &quot; 10 #3 C)2
U(t = \$) =
=
= 25mJ
#6
2 C 2 20 &quot; 10 F
Q(t = 1.25ms) = Q(0)e#t / RC = 1.0 &quot; 10 #3 e#1.25 ms /1ms = 0.2865 &quot; 10 #3 C
1 Q 2 1 (0.2865 &quot; 10 #3 C)2
U(t = 1.25ms) =
=
= 2.05mJ
#6
2 C 2
20 &quot; 10 F
%ER = U(t = \$) # U(t = 1.25ms) = 23mJ
```