Kirchhoff’ second rule
Close a battery on a resistor: simplest circuit!
I
+
-
V
R
How much current flows in the circuit? Ohm’s law: I =
V
R
When the current flows in a resistor there is a voltage drop ∆V=-IR
Kirchhoff’s second law:
Around any closed loops, the sum of EMF and potential drops is 0
Equivalent to say that Electrostatic filed is conservative:
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∫ E ids = 0
8.022 – Lecture 8
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Solving circuits
If we have more than 1 resistor:
V
R1
I
+
-
R2
R3
Solve the circuit: determine currents and voltages everywhere
What we know:
Current flowing in the circuit must be the same everywhere, or Q would
accumulate somewhere
Voltage drop in the ith resistor: ∆Vi=-IRi
Second Kirchhoff rule: V − ∑ Vi = 0
i
V − ∑ Vi = V − I ∑ Ri = 0 ⇒ I =
i =1,3
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i =1,3
V
R1 + R 2 + R3
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5
Resistors in series
We implicitly derived an important result. We wrote:
V − ∑ Vi = V − I ∑ Ri = 0 ⇒ I =
i =1,3
i =1,3
V
R1 + R2 + R3
What does it mean? Same current flowing in these two circuits:
R1
V
I
+
-
R3
R2
≡
V
+
-
I
R eq = ∑ Ri
i
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Req
8.022 – Lecture 8
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Solving circuits
Solve the circuit: determine currents and voltages everywhere
V
A
R1
+
-
I
D
R3
B
R2
C
Calculate VAB, VBC, VCD, VDA, VAC, …
VAB = IR1 =
VR1
VR2
; VBC = IR2 =
R1 + R2 + R3
R1 + R2 + R3
VCD = IR3 =
VR3
V ( R1 + R2 )
; VAC = I ( R2 + R1 ) =
R1 + R2 + R3
R1 + R2 + R3
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8.022 – Lecture 8
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6
Application (F13)
What is the resistance of electrical components?
Elements of the circuit:
R5
R4
- Saline solution
- Resistor
- Diod
- light bulb
-Fluoreschent light
R3
R1
R2
How to measure knowing the current = 135 mV?
You are given a voltmeter!
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8.022 – Lecture 8
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Kirchhoff’s first rule
Let’s now connect resistors in parallel:
I
N
V
I1
R1
I2
R2
At the node N the current I divides up into 2 pieces: I1 and I2
Kirchhoff’s first law:
At any node, sum of the currents in = sum of the currents out
In other words: there is no accumulation of charges in the circuit
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8.022 – Lecture 8
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7
Kirchhoff’s first rule: application
Solve the circuit:
V = I1 R1
⇒
I1 =
V = I 2 R2
⇒
I2 =
I
V
R1
I1
V
V
R2
N
R1
I2
R2
Apply Kirchhoff’s first law: I=I1+I2
⎛1
1 ⎞
I = I1 + I 2 = V ⎜ + ⎟
⎝ R1 R2 ⎠
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8.022 – Lecture 8
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Resistors in parallel
Again, we are learning something important. We said:
⎛ 1 1 ⎞
I = I1 + I 2 = V ⎜ + ⎟
⎝ R1 R2 ⎠
What does it mean?
I
V
I1
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R1
I2
R2
≡
V
+
-
1
1
=∑
R eq
i Ri
8.022 – Lecture 8
I
Req
16
8
Resistors in parallel vs. in series
Resistors in series:
The current flowing is one
add resistors
make the path harder
I decreases
Req increases
Req larger than any single resistor
R eq = ∑ Ri
I
i
Resistors in parallel:
The current flows is many resistors
add resistors
make path easier
I increases
Req is smaller than any single resistor
1
1
=∑
R eq
i Ri
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8.022 – Lecture 8
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Application (F12)
Consider the two circuits:
V= 1.5 V
RL
V
R L = 1580 Ω
Ammeter reading: 0.94 mA
A
R
R
R L = 1580 Ω
V
R
RL
R = 912 Ω
Ammeter reading: ?? mA
A
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9
Slightly harder circuits
How do we solve this?
R1
V1
I1
R2
Reducing the circuit does not work:
I2
V2
R3
Series and parallels won’t work
Because of second EMF
But Kirchhoff still holds so:
Apply First Kirchhoff law to each node
Apply Second Kirchhoff law to each loop
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8.022 – Lecture 8
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Slightly harder circuits (2)
R1
Solution:
Left loop:
V1-I1R1-(I1-I2)R2=0
Right loop:
-V2-I2R3-(I2-I1)R2=0
Node:
I(in R2)=I1-I2
V1
R2
I2
V2
R3
Solving the system:
V 1R3 + (V 1 − V 2) R 2
I1 =
R1R 2 + R1R3 + R 2 R3
(V 1 − V 2) R 2 + V 2 R1
I2 =
R1R 2 + R1R3 + R 2 R3
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I1
How to go through a loop?
• Assign current direction (arbitrary)
• Choose a path (clockwise or ACW)
• EMF: >0 when - + and <0 when + • V always drops on R
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10
Internal resistance
When battery delivers current to circuit there is a flow of current in
the battery itself.
In previous example this current comes from flux of H+ ions
The chemical reaction will dissipate energy, battery gets hot, energy
that could have gone in the circuit is lost
This is equivalent to having a resistor r inside the battery:
V
r
Corollary:
There is a max current the battery can generate: Imax=V/r
Homework:
Short a little battery with a wire and see how hot it gets: careful! ☺
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8.022 – Lecture 8
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Power dissipated in resistors
When current flows in circuit, it moves charges through ∆V
work
Work done to drive a charge dq through a potential difference V:
dW = Vdq
If work is done in time dt
the power dissipated is:
dW
dq
=V
= VI
P=
dt
dt
When Ohm’s law hold:
P = VI = RI 2
Units: [P]=[Energy]/[time]
cgs: erg/s
SI: J/s
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Power is important: it’s what does work in a circuit:
how much light is produced, how much heath, etc.
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11
Dependence of R on T (F16)
Ohm’s law tells us that V=RI
This is valid in any resistor
Does it mean that given a voltage I is constant over time?
Not necessarity!
When I goes through R it dissipates power=RI2
R=f(T)
R is not constant while resistor heats up! Last time: increase T
increase R
decrease I
Pulse a light bulb with a saw tooth voltage:
What do you expect to observe?
As the filament warms up, R changes
the slope of I vs V changes over time
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Summary and outlook
Today:
First look at circuits: Kirchhoff’s laws
Resistors in series and parallel
Power dissipated in resistors
Next time:
RC circuits and evolution of currents over time
Remember: come and talk to us today if test < 50
Make appointment by email is safer
Add date is THIS FRI!!!
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12