Physics 227: Lecture 11
Circuits, KVL, KCL, Meters
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Lecture 10 review:
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EMF ξ is not a voltage V, but OK for now.
Physical emf source has Vab = ξ - Irinternal.
Power in a circuit element is P = IV.
For a resistor with V = IR, P = IV = I2R = V2/R.
Light bulbs are rated in Watts, for fixed V over them.
Higher power bulbs have lower resistance, do brightness of
bulbs in series is opposite brightness of bulbs in parallel.
Monday, October 17, 2011
Node
Definitions
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A node or junction: a place where 2 or
more wires meet. Often shown in drawings
as a solid circle.
Elements (resistors, capacitors, ...) in
parallel: elements between the same two
notes, necessarily have the same voltage
over them.
Elements in series: elements connected
with a wire, with no nodes between them,
so that they necessarily have the same
current through them.
E1 and E2 in parallel
E2
E1
E1 and E2 in series
E1
Monday, October 17, 2011
E2
Resistor iClicker
Which of the following is true
about the circuit shown?
A. R1 is in parallel with R2.
B. R1 is in parallel with R3.
C. R2 is in parallel with R3.
D. R1 is in series with R2 and/or R3.
E. R2 is in series with R3.
Monday, October 17, 2011
Resistor iClicker
Which of the following is true
about the circuit shown?
A. R1 is in parallel with R2.
B. R1 is in parallel with R3.
C. R2 is in parallel with R3.
D. R1 is in series with R2 and/or R3.
E. R2 is in series with R3.
Monday, October 17, 2011
R2 and R3 are in series they are connect with a
wire and have the same
current. No other
resistors are in series.
No resistors are in
parallel - no two connect
the same two nodes and
necessarily have the
same voltage.
KVL and KCL
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I1
Two basic ideas for analyzing circuits:
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No charge buildup in a circuit:
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If wires meet, the sum of the
currents into (or out of) a node is 0.
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I2
∑i Ii = 0.
``Kirchoff’s Current Law’’
If you go around a loop, the voltage is
the same.
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I3
You return to a position with the
same potential and potential energy.
∑i Vi = 0.
``Kirchoff’s Voltage Law’’
Monday, October 17, 2011
V4
V1
V3
V2
Resistors in series
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KCL: there is the same current I through each of the resistors.
V = V1 + V2 + V3 = I(R1 + R2 + R3).
e.g.: V1 / V = [ R1 / (R1 + R2 + R3) ].
We can replace several resistors in series with one equivalent
resistor: Req = R1 + R2 + R3 + ...
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Resistors in series add. This is opposite the case for capacitors.
Monday, October 17, 2011
Resistors in parallel
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KVL: there is the same voltage V across each of the resistors.
The total current I is split between the resistors.
I = I1 + I2 + I3 = V/R1 + V/R2 + V/R3 = V(1/R1 + 1/R2 + 1/R3).
e.g.: I1 / I = [ 1/R1 / (1/R1 + 1/R2 + 1/R3) ] =
(R2R3) / (R1R2 + R2R3 + R1R3)
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We can replace several resistors in parallel with one equivalent
resistor: 1/Req = 1/R1 + 1/R2 + 1/R3 + ...
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Resistors in parallel add inversely. This is opposite the case for
capacitors.
Monday, October 17, 2011
Simple Networks - series / parallel
Monday, October 17, 2011
Simple Networks - series / parallel
Monday, October 17, 2011
Simple Networks - series / parallel
Monday, October 17, 2011
Simple Networks - series / parallel
Monday, October 17, 2011
Less Simple Networks - applying KVL/KCL
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This circuit cannot be simplified
using parallel / series, we need
to use KVL / KCL.
Define the currents going
around loops - do not worry
about the actual current
direction, if you guess wrong
you will get negative numbers.
• Note the current through an
Sometimes they will be solved as
n equations in n unknowns, or
perhaps we can more ``artfully’’
find easy quantities first.
Monday, October 17, 2011
element may be the sum of two
loop currents. E.g.:
Ithrough r2 = Iloop2 - Iloop3
• As shown, only two of I
loop1,
Iloop2, and Iloop3 are independent.
Cicuitous iClicker
A. I is impossible to figure out.
What is I?
B. I = 0 A.
C. I = 0.5 A.
D. I = (8/9) A.
E. I = (8/3) A.
Monday, October 17, 2011
Cicuitous iClicker
What is I?
A. I is impossible to figure out.
Apply KVL going CCW
around the circuit from b...
B. I = 0 A.
12-2I-3I-4I-4-7I = 0
8-16I = 0
I = 1/2
Monday, October 17, 2011
C. I = 0.5 A.
D. I = (8/9) A.
E. I = (8/3) A.
Less Simple Networks - applying KVL/KCL
• What is I? Use KCL at node a:
• I=1A+2A=3A
• What is r? Use KVL around loop 1:
• -(2 A)(3 Ω) + 12 V - (3 A)r = 0 ➭ r = 2 Ω
• What is ξ? Use KVL around loop 2:
• -(2 A)(3 Ω) - ξ + (1 A)(1 Ω) = 0 ➭ ξ = -5 V
Monday, October 17, 2011
Voltmeters and Ammeters
• Old electro-mechanical systems largely based
on forces between current carrying wires and
magnetic fields - a topic for the near future.
• Modern system all ICs.
Monday, October 17, 2011
Voltmeters and Ammeters
• For a voltmeter, you want a
large internal meter
resistance. Any current
flowing through the meter
does not flow through the
circuit element, and changes
what is happening in the
circuit.
Rest of circuit
If Rvoltmeter is made too small,
most of current flows through it,
and there is a greater voltage in Rvoltmeter
the rest of the circuit. Similarly,
for ammeter you want R small.
Monday, October 17, 2011
R to measure
Voltage over
Voltmeters and Ammeters
• Ammeters and Voltmeters have a dial to turn. This allows us to
select the ratio of Rc to Rs, varying the current through the
meter, so that we can change the magnitude of voltage / current
that gives a full scale reading.
• E.g.: too much current, needle pegged at full scale, increase R to
reduce current. Too little current, reading insensitive due to small
needle deflection, decrease R to get more current and more
needle deflection.
Monday, October 17, 2011
Voltmeters and Ammeters
• Which is the right way to measure the resistance of a resistor?
Monday, October 17, 2011
Voltmeters and Ammeters
• Use an ohmmeter so you do not have any corrections.
Monday, October 17, 2011
Current iClicker
A constant current source sends a current I through a resistor R.
A second resistor R’ (<< R) is connected in parallel to the first resistor.
What happens to the current I through the first resistor?
A. I increases.
B. I decreases, but not to 0.
C. I decreases to 0.
D. I stays the same.
E. I cannot figure out what happens.
Monday, October 17, 2011
Current iClicker
A constant current source sends a current I through a resistor R.
A second resistor R’ (<< R) is connected in parallel to the first resistor.
What happens to the current I through the first resistor?
R
R’
A. I increases.
B. I decreases, but not to 0.
C. I decreases to 0.
With just R, we have V=I0R.
Resistors in parallel act as a
current divider. The current
D. I stays the same.
will split between the two so
E. I cannot figure out what happens. that V = IR = I’R’, and I+I’ =
I0, or V (1/R + 1/R’) = I0.
Monday, October 17, 2011
Thank you.
See you monday.
Monday, October 17, 2011