Electric circuit components
Capacitor – stores charge and potential
energy, measured in Farads (F)
Battery – generates a constant electrical
potential difference (∆V) across it.
Measured in Volts (V).
Resistor – resists flow of charge due to
scattering; dissipates energy.
Measured in Ohms (Ω)
Direct Current (DC) circuits
First, we will consider circuits with batteries and resistors.
V=5V
I
R
Resistors in Series
The current through all
resistors is the same
I1 = I 2 = I 3 = I
V1 = Vax = IR1
V2 = Vxy = IR2
V3 = V yb = IR3
Vab = V1 + V2 + V3 = I (R1 + R2 + R3 )
Vab = IReq → Req = R1 + R2 + R3
Resistors in Series
Req = ∑ Ri
i
Adding resistors in series always increases the total resistance.
Resistors in Series: Example
Calculate I and V across each resistor
Ibat
RA
IA
RA = 1.0 Ω
IB
RB = 3.0 Ω
V
V = 6 Volts
RB
Clicker Question
Two different wires are connected to a battery in
series (in a chain, one after another). How does
the current in upper wire A compare to the current
in lower wire B?
V
A) iA > iB
B) iA < iB
C) iA = iB
D) Impossible to determine.
Ibat
RA
IA
RB
IB
Resistors in Series: Example
What is the equivalent resistance?
Req = R1 + R2 = 1.0Ω + 3.0Ω = 4.0Ω
Ibat
What is the current through the circuit?
Assuming everything is Ohmic, V=IR
i = i A = iB =
RA
IA
RB
IB
V
V
6V
=
= 1.5 A
Req 4.0Ω
Resistors in Series: Example
What about the Voltage change across each resistor?
Ibat
RA
IA
RA = 1.0 Ω, iA=1.5 A
IB
RB = 3.0 Ω, iB = 1.5 A
V
V = 6 Volts
RB
V A = ∆VA = i A RA = (1.5 A)(1.0Ω) = 1.5V
VB = ∆VB = iB RB = (1.5 A)(3.0Ω) = 4.5V
Clicker Question
If we consider the Voltage change starting
Ibat
at Point P and going all the way around the
circuit loop back to Point P, what is the total V
∆V?
Α)∆V = +6.0 Volts
Β)∆V = +1.5 Volts
C) ∆V = -4.5 Volts
D) ∆V = 0.0 Volts
Point P
E) ∆V = -6.0 Volts
IA
RA
RB
IB
Resistors in Series: Example
Voltage
P2
P3
Ibat
RA
IA
6V
4.5V
V
P4
RB
P1
IB
P5
Define V=0 at P1.
3.0V
1.5V
0.0V
1
2
3
4
5
1 Position
In Loop
Clicker Question
We start with the left circuit with one lightbulb (A). The
brightness of the bulb directly reflects the power. If we add a
second bulb (B) as shown on the right, what happens to the
bulbs?
A) Bulb A is equally
bright.
B) Bulb A is dimmer
V
V
than before
A
A
B
C)Bulb A is brighter
than before
Clicker Question
Two light bulbs, A and B, are in series, so they carry the
same current. Light bulb A is brighter than B.
Which bulb has higher resistance?
A) A
B) B
C) Same resistance.
Answer: Bulb A has higher resistance.
Since the resistors are in series, they
have the same current I. According to
P = I2 R, if I = constant, then higher R
gives higher P.
V
A
B
Clicker Question
If we connect a battery with
V = 6 Volts with three resistors
R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω,
which of the following is true?
A) The current through each resistor
is the same.
B) The Voltage change through
each resistor is the same.
C)The resistance of each resistor is
the same.
D)None of the above are true.
Resistors in Parallel
The voltage across all
resistors is the same
V1 = V2 = V3 = Vab = V
I1 = V / R1
I 2 = V / R2
I 3 = V / R3
1 1
1 ⎞
+
+ ⎟⎟
R3 ⎠
⎝ R1 R2
⎛
I = I1 + I 2 + I 3 = V ⎜⎜
I = V / Req →
1
Req
=
1
R1
+
1
R2
+
1
R3
Resistors in Parallel
1
1
=∑
Req
i Ri
Adding resistors in parallel always decreases the total resistance.
Traffic analog
Resistors in
parallel are like
having more lanes
on the highway.
This reduces the
resistance for
getting from one
place to another.
Resistors in parallel: Example
i1
i
i2
R1
R2
I = I1 + I 2 + I 3
R3
i3
V
If all three resistors were the
same then:
1
I1 = I 2 = I 3 = I
3
Resistor configurations
Resistors in Series
Req = R1 + R2 + R3
Always increases the resistance!
Resistors in Parallel
Req =
1
1
1
1
+
+
R1 R2 R3
Always decreases the resistance!
Clicker Question
We start with the left circuit with one lightbulb (A). The
brightness of the bulb directly reflects the power. If we add a
second identical bulb (B) as shown on the right, what
happens to the bulbs?
B
A
V=12V
Clicker Question
A) Bulb A is equally
bright.
B) Bulb A is dimmer
than before
C)Bulb A is brighter
than before
A
V=12V
If each of these six light bulbs is
identical, which bulb is going to
be the brightest?
A) Bulb A
B) Bulb B
C)Bulb C
D)Bulb D
E) Bulb E
Clicker Question
The three light bulbs A, B, and C are identical. How does the
brightness of bulbs B and C together compare with the
brightness of bulb A?
A) Total power in B+C = power in A.
A
B) Total power in B+C > power in A.
C) Total power in B+C < power in A.
B
C
V=12V
Answer: Use P = V2/Rtot For bulbs B
and C, Rtot = 2R.
Total power in B+C < power in A.
Clicker Question
In the circuit below, what happens to the brightness of bulb 1,
when bulb 2 burns out? (When a bulb burns out, its resistance
becomes infinite.)
2
A) Bulb 1 gets brighter
B) Bulb 1 gets dimmer.
C) It's brightness remains the same.
3
1
(Hint: What happens to the current from
the battery when bulb 2 burns out.)
V=12V
Answers: Bulb 1 gets dimmer! When bulb 2 burns outs, the
filament inside breaks and R2 becomes infinitely large. The
total equivalent resistance which the battery sees increases
(since bulb 2 is gone, there are fewer paths for the current
flow, so less flow, more total resistance.) Since the battery
sees a larger R-tot, the current from the battery I-tot = V/Rtot
is reduced. Less current from the battery means less current
2
through bulb 1, less light.
3
1
V=12V
CPS question
Which is the best way to wire a house?
(A)
(B)
Resistor networks
Resistor networks
Kirchhoff’s rules
Kirchhoff’s Voltage Loop Law:
The change in Voltage around any closed loop must be zero.
Kirchhoff’s Current Junction Law:
In steady state, the current going into
a junction (or point) must
equal the current going out of that
junction (or point).
Clicker Question
What is the electric potential difference across the upper light
bulb (resistor)? Think about our Voltage Loop Rule.
A) |V| = 0 Volts
B) |V| = 6 Volts
C)|V| = 12 Volts
D)|V| = 24 Volts
E) None of the above answers.
V=12V
Kirchhoff’s rules: Example
Example 1: Choose R1 = R2 = R3 = 10 Ω
1. Longer line on battery is
higher voltage and the
shorter line is lower voltage,
i1
by convention.
2. We label the current in
1
each section as shown.
i
b
Note that we can guess the
direction at this point. The
answer will be independent
of our guess.
ib
2 i2
i3
3
+
ib
ib
V=12V
Kirchhoff’s rules: Example
3. Apply the Voltage Loop
Rule around each possible
loop in the circuit.
2 i2
Make sure to label the
direction of your loop!
Make sure to pick a starting
point to go around the loop. ib
Then add up the Voltage
changes around the loop.
i1
i3
3
1
ib
+
ib
V=12V
ib
Critical Sign Convention!
-
+
V=12V
-
+
V=12V
If your loop goes through a battery from – to +
the Voltage increases (e.g. ∆V = +12 V)
If your loop goes through a battery from + to –
the Voltage decreases (e.g. ∆V = -12 V)
Critical Sign Convention!
i
2
i
2
If you go across a resistor and the loop
direction and guessed current direction are
the same, the voltage decreases (e.g. ∆V = -iR)
If you go across a resistor and the loop
direction and guessed current direction are
opposite, the voltage increases (e.g. ∆V = +iR)
Kirchhoff’s rules: Example
Starting at Point P go around the loop.
2 i2
∆Vbat = +12V
∆VR 3 = −i3 R3
∆VR1 = −i1 R1
i1
ib 0
∆Vsum = +12 − i3 R3 − i1 R1 =
P
i3
3
1
-
+
ib
ib
V=12V
ib
Kirchhoff’s rules: Example
Now try other path starting at Point P around the new loop.
2 i2
∆Vbat = +12V
∆VR 2 = −i2 R2
∆VR1 = −i1 R1
i1
∆Vsum = +12 − i2 R2 − i1 R1 = 0
P
i3
3
1
ib
+
ib
V=12V
ib
Kirchhoff’s rules: Example
For our specific circuit, we used R1 = R2 = R3 = 10 Ω
∆Vsum = +12 − i3 R3 − i1 R1 = 0
12 − 10i3 − 10i1 = 0
∆Vsum = +12 − i2 R2 − i1 R1 = 0
12 − 10i2 − 10i1 = 0
We have two equations and three unknowns.
We need additional information.
Kirchhoff’s rules: Example
Apply the current junction rule at point Q below.
iin = iout
2 i2
i2 + i3 = i1
Also, (as we could have
done at the start):
i1 = ib
i1
ib
1
i3
Q
ib
3
+
ib
V=12V
ib
Kirchhoff’s rules: Example
12 − 10i3 − 10i1 = 0
12 − 10i2 − 10i1 = 0
i2 + i3 = i1
i1 = 0.8 A
Solve 3 equations and 3 unknowns.
i 2 = 0 .4 A
i3 = 0.4 A
Kirchhoff’s rules: Example
Does this make sense?
i2 = i3 since they are two parallel,
equal resistance paths.
Current from R2 and R3 must go
into R1.
i1 = 0 .8 A
i 2 = 0 .4 A
i3 = 0.4 A
i1
2 i2
i1
i3
3
1
i1
+
i1
V=12V
i1
Clicker Question
R1 = 10 Ω i1 = 0.8 Amps
R2 = 10 Ω i2 = 0.4 Amps
What is the value of the
equivalent resistor that R3 = 10 Ω i3 = 0.4 Amps
i1
would replace the three
resistors below?
A) Req = 5 Ω
1
B) Req = 10 Ω
i1
C)Req = 15 Ω
D)Req = 30 Ω
E) None of the Above
i1
1
⎛ 1
⎜
⎜R
⎝ 2
1 ⎞
⎟
+
R 3 ⎟⎠
3
+
i1
V=12V
2 i2
i3
3
1
Then R23 and R1 are in series.
i1
i1
i1
i1
= 5Ω
Req = R123 = R1 + R23 = 15Ω
i3
R1 = 10 Ω i1 = 0.8 Amps
R2 = 10 Ω i2 = 0.4 Amps
R3 = 10 Ω i3 = 0 .4 Amps
Two ways to think about it.
1. Battery of 12 V has 0.8 Amps
coming out of it. By V=iR the
Req = 12V/0.8A=15 Ω
2. R2 and R3 are in parallel.
R23 =
2 i2
+
i1
V=12V
i1
Clicker Question
Which of these diagrams represent the same circuit?
A. a and b
B. a and c
C. b and c
D. a, b, and c
E. a, b, and d
Clicker Question
A circuit with two batteries is shown below. The directions of
the currents have been chosen (guessed) as shown.
Which is the correct current equation for this circuit?
A) I2 = I1 + I3
B) I1 = I2 + I3
C) I3 = I1 + I2
I1
D) None of these.
R1
V1
R2
R3
I3
I2 Loop 1.
V2
Clicker Question
Which equation below is the correct equation for Loop 1?
A) –V2 + I1R1 – I2R2 = 0
B) V2 + I1R1 – I2R2 = 0
C) –V2 – I1R1 + I2R2 = 0
D) V2 + I1R1 + I2R2 = 0
E) None of these.
I1
I3
Answer: –V2 + I1R1 – I2R2 = 0
R1
V1
R2
I2 Loop 1.
R3
I3
Finish a full analysis of this circuit. Consider case:
V1 = 12 V, V2 = 24 V, R1=R2=R3 = 1 Ω.
I1
I3
R1
V1
R2
Loop 2
I2 Loop 1.
V2
R3
I3
i1 = 20 A, i2 = −4 A, i3 = 16 A
i3 = i1 + i2
− V2 + i1 R1 − i2 R2 = 0
− 24 + i1 − i2 = 0
+ V1 − i2 R2 − i3 R3 = 0
+ 12 − i2 − i3 = 0
V2
Clicker Question
Two light bulbs A and B are connected in
series to a constant voltage source. When a
wire is connected across B as shown, the
brightness of bulb A ...
A: increases
B: decreases, but remains glowing
C: decreases to zero (bulb A goes
completely dark, no current)
D: remains unchanged
Answer: Bulb A increases in brightness
(since the current increases when B is
shorted).
Circuit Probes
An Ammeter measures the current through itself.
An ideal Ammeter has zero resistance so it does not affect the
circuit!
i
i
A
To measure the current
through R, must place
Ammeter in series.
A
V=5V
I
R
Clicker Question
Consider the circuit shown. If you want to measure the current
thru bulb 3, how should the ammeter be attached (assume you
only attach one at a time)?
A) Pink
Pink
1
B) Yellow
A
A
Blue
A
C) Green
4
3
Yellow 2
V
D) Blue
A
Green
E) Two of the 4 positions
can be used to measure
bulb3’s current.
Answer: B) Yellow
Circuit Probes
To measure voltage across R, must place Voltmeter in
parallel with R.
You want the Voltmeter to be very high resistance (ideally
infinity) to avoid drawing current.
V=5V
I
R
V
Clicker Question
In the circuit shown, what does the voltmeter read?
A:
B:
C:
D:
E:
6V
3V
2V
0V
Voltmeter will "fry"
6V
2Ω
V
Clicker Question
Now you switch the voltmeter over to "amp" mode. (But you
leave it in the same position in the circuit) What does the
Ammeter read?
A:
6A
B:
3A
6V
A
2Ω
C:
2A
D:
0A
E:
Ammeter will "fry"
Ammeter will fry. Ideal Ammeter has zero internal resistance. If you attach an
ideal ammeter to a battery, you will get infinite power.