Direct Current (DC) Circuits
Circuit Basics
wire
open switch
closed switch
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
2-way switch
1.5 V
+
–
47 F
4.7 k
ideal battery
capacitor
resistor
•Two devices are in parallel if they are
connected at both ends
Resistors in Parallel and in Series
R1
V1  IR1
R2
V2  IR2
•When resistors are in series, the same
current must go through both of them
•The total voltage difference is
V  V1  V2  I  R1  R2 
•The two resistors act like one with
resistance
R  R1  R2
V  I1 R1
V  I 2 R2
R1
R2
•When resistors are in parallel, the
same potential is across both of them
•The total current through them is
V
 I  I1  I 2  V  V
R
R1
R2
•The two resistors act like one with
resistance
1 1
1
 
R R1 R2
Parallel and Series - Formulas
Capacitor
Resistor
Inductor*
1 1
1
 
C C1 C2
R  R1  R2
L  L1  L2
Parallel
C  C1  C2
1 1
1
 
R R1 R2
1 1 1
 
L L1 L2
Fundamental
Formula
Q
V 
C
V  IR
dI
EL   L
dt
Series
* 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
R1
V1  IR1 
E
R1  R2
V2  IR2 
R2
E
R1  R2
R  R1  R2
I
E
R1  R2
The larger resistor gets
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
Ideal vs. Non-Ideal Batteries
realistic battery
10  30 V
–
–
E
ideal battery
50 
+
–
R1
V1 
E  25 V
R1  R2
r
+
A 30 V battery with 10  of internal resistance
is connected to a 50  resistor. What is the
actual voltage across the 50  resistor?
E
+
•Up until now, we’ve treated a battery as if it
produced a fixed voltage, no matter what we
demand of it
•Real batteries also have resistance
•It limits the current and therefore the
power that can be delivered
•If the internal resistance r is small compared
to other resistances in the problem, we can
ignore it
Kirchoff’s Laws
Kirchoff’s Second Law:
The total voltage change
around a loop is always zero
I
in
  I out
•These yield a series of equations, which you then solve
I1
3
Kirchoff’s First Law:
The total current into any
vertex equals the current
out of that vertex
+
I2
+
•Kirchoff’s Laws help us figure out where and how
much current is flowing in a circuit
•The first step is to assign a direction and a current to
every part of a circuit
•Items in series must have the same current in them
•Then you apply the two laws, which can be thought of
as conservation of charge and conservation of voltage,
which you apply to vertices and loops respectively.
12 V
–
–
5
6V
4
I3
Kirchoff’s First Law
12 V
–
+
You always get one
redundant equation
B
– A
6V
+
I2
•A vertex is any place where three or more wires
come together
•For example, at point A, this gives the equation:
I1  I 2  I 3
•At point B, this gives the equation:
I 3  I1  I 2
I1
3
Kirchoff’s First Law:
The total current into any
vertex equals the current
out of that vertex
5
4
I3
Kirchoff’s Second Law
0  18  3I1  5I 2
I1
3
0  12 3I1 6 5I 2
12 V
–
+
I2
+
Kirchoff’s Second Law:
•First, pick a direction for every
The total voltage change
loop
around a loop is always zero
•I always 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
+
12 V
–
 Iin   Iout
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
You always get one
redundant equation
3
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:
I1
B
– A
6V
+
I2
5
4
I3
Kirchoff’s Law- Final Step
0  5I 2  6  4 I3
•We now solve these simultaneously
•We can let Maple do it for us if we want:
> solve({i3=i1+i2,0=-5*i2-6.-4*i3,
0=18-3*i1+5*i2},[i1,i2,i3]);
I2
5
4
•Negative currents mean we guessed the wrong way
•Not a problem
– A
6V
+
0  18  3I1  5I 2
I1
3
I 3  I1  I 2
+
•You have derived three equations in three unknowns
12 V
–
I3
Kirchoff’s Laws with Capacitors
•Pick one side to put the charge on
•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 a minus sign if I isn’t on the same side as Q
•If you are in a steady state, the current through a
capacitor is always zero
Q
C
+
–
dQ
I
dt
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
R
Q0
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
dQ
Q
0   RI
 I  
C
dt
RC
•This is a differential equation, and therefore hard to solve
dQ
dt

Q
RC
dQ
dt
 Q   RC
t
ln Q  
k
RC
Q  et RC
Q  Q0 e
t RC
Check the units:
RC  F
V C

A V
C
Cs
s
Charging and Discharging Capacitors
Q  Q0 e t RC
•The combination RC =  is called the time constant
•It’s the characteristic time it takes to discharge
t 
Q

Q
e
0
•We can work out the current from
dQ
 Q0 et 
I 
dt
In this circuit, the
Q
capacitor is initially
uncharged, but at t = 0
the switch is closed. C
What happens?
dQ
I
dt
Q
0   IR  E
C
I
R
+
–
E
dQ
Q E


dt
RC R
Q  EC 1  e t RC 
  RC
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
Household Wiring
*Actually, this
•All household appliances consume electrical power
is alternating
•Think of them as resistors with fixed resistance R
current, later
•Devices are designed to operate at 120 V*
2
chapter
•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)
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