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Transcript
Examining Voltage Drops
Around a Closed Loop
+
5V
-
220W
470W
ENGR 120
Select Resistors
Find the 220W and the 470W resistors from your Boe-Bot kit.
Example: 470W resistor:
4 = yellow
7 = violet
Add 1 zero to 47 to make 470, so 1 = brown
So, 470 = yellow, violet, brown
Now, find the 220W resistor.
Build the Series Circuit Below
+
5V
-
220W
470W
the breadboard connects
these resistors
Compute the Voltage Drops Across the Two Resistors
(Hint: compute the total current provided by the power source
and then apply Ohm’s Law to each resistor)
5V
+ 220W
DV =
470W
DV =
SOLUTION:
Now, add the voltage rise of
the power source (+5V) to the
voltage drops across the resistors
(negative numbers).
V–
V–
V=
V
Use Multimeter to Measure Voltages Around Loop
(1) From Vdd to Vss
DV1 = _____
(2) Across the 470W resistor
DV2 = _____
Remember . . .
a RESISTOR is a voltage DROP and
a POWER SOURCE is a voltage RISE
DV1 - DV2 - DV3 = ________
(3) Across the 220W resistor
DV3 = _____
Rises must balance drops!!!!
Compare Measurements to Theory
5V
+
-
+
DV =
470W
-
220W
+
-
DV =
How close are your
experimental and
theoretical results?
Kirchoff’s Voltage Law (KVL)
V–
+
V–
-
+
Kirchoff’s Voltage Law says that the algebraic
sum of voltages around any closed loop in a
circuit is zero – we see that this is true for our
circuit. It is also true for very complex circuits.
5V
DV =
470W
-
V=
V
220W
+
-
DV =
Notice that the 5V is DIVIDED between the two
resistors, with the larger voltage drop occurring
across the larger resistor.
Gustav Kirchoff (1824 – 1887) was a German physicist who made fundamental contributions
to the understanding of electrical circuits and to the science of emission spectroscopy. He
showed that when elements were heated to incandescence, they produce a characteristic
signature allowing them to be identified. He wrote the laws for closed electric circuits in 1845
when he was a 21 year-old student.
Photo: Library of Congress
The End 