Download DC Circuits

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Spark-gap transmitter wikipedia , lookup

Electrical ballast wikipedia , lookup

Switched-mode power supply wikipedia , lookup

Voltage optimisation wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Opto-isolator wikipedia , lookup

TRIAC wikipedia , lookup

Distribution management system wikipedia , lookup

Earthing system wikipedia , lookup

Current source wikipedia , lookup

Mains electricity wikipedia , lookup

Surge protector wikipedia , lookup

Stray voltage wikipedia , lookup

Two-port network wikipedia , lookup

Buck converter wikipedia , lookup

Metadyne wikipedia , lookup

Rectiverter wikipedia , lookup

Ohm's law wikipedia , lookup

Alternating current wikipedia , lookup

Network analysis (electrical circuits) wikipedia , lookup

Transcript
Revised 8/03
Experiment 22: DC Circuits - Kirchoff’s Laws
Purpose:
To study the principles of DC networks and the use of Kirchoff’s Rules.
Apparatus
1) a pre-wired circuit board with batteries, DC millimeter
2) a DC voltmeter electrical connectors
Theory: Definitions
1) A junction is a point where two or more leads connect in an electrical circuit.
2) A branch connects two neighboring junctions. Each branch carries its own branch current.
3) A loop is a closed path, which starts and ends at the same junction.
Theory
A) Kirchoff’s Current Law: The algebraic sum of all branch currents at a junction equals zero.
Σ Ii = 0
(1)
[Sign Convention: all currents entering the junction are assumed positive (+I), while all those
leaving it are assumed negative (-I).]
So, e.g., at a junction with three currents, I1 enters, while I2 and I3 leave, would be written, using
Kirchoff's current law as: +I1 - I2 - I3 = 0. From which one may deduce that: I1 = I2 + I3. This
makes sense since I1 is supplying current to the junction while I2 and I3 are removing current
from the junction, or, ΣIin = ΣIout. (See Fig. 1.)

It makes no difference whether this is the correct direction or not. You will find the correct
direction after you solve the equations because solutions with the negative sign imply that the
actual direction of the current is opposite to the assumed direction.
111
Experiment 22
B) Kirchoff’s Voltage Law: The algebraic sum of all potential differences in a loop is equal to
zero.
Σ Vi = 0
(2)
[Sign Convention: The side of a resistor, capacitor, or inductor (passive circuit component) where
the current enters is positive (+V) since the current loses energy as it passes through the
component and the entry side is at a higher potential. The other side is negative (-V) since it is at
a lower potential. The side of a battery or power supply (active circuit component) where the
current enters is negative and the other side is positive (meaning that the current gains energy as
it passes through the component.]
There are two types of potential differences in any circuit: 1) a voltage difference, (designated by
a "V") and 2) an electromotive force (emf or , aka batteries or power supplies) designated by
“+/- ”.
To set up an equation like (2), trace a path around a closed loop. Then, when you pass through a
battery write + if your path gains energy. Passing through a resistor, capacitor, or inductor, write
the voltage as "-V" if you are losing energy (same direction as the current). Use labels
appropriately.
For example, look at the simple series circuit with three voltages: , V1, and V2 shown in Figure
2. The current direction is clockwise. The sign convention above gives us the positive and
negative signs for the 2 resistors and the battery in the circuit as shown. Going around the loop
in the clockwise direction, there is a voltage gain through the battery and voltage losses through
the resistors. So in this direction, Kirchoff's voltage law gives us + - V1 - V2 = 0. From which
one may deduce that  = V1 + V2. The two resistors (V1, and V2) use up all of the voltage
supplied by the battery (). (See Fig. 2.)
V1
-
+
I
 +
I
+
V2
-
I
I
Fig. 2 Kirchoff's Voltage Law
112
Experiment 22
C) Ohm's law: This is the basic form of all voltage drops. The current, I, passes through the
resistance, R, which requires a potential difference of V to push the electrons through it. The
equation is one of the simplest in all of physics:
V=I·R
(3)
Preliminary Procedure
a) Your network board is shown in Fig. 3. Familiarize yourself with it and make sure that the
batteries are mounted with polarities conforming to Fig. 3.
a
1
c
b
e
d
f
R1
R3
k
R2
2
R4
j
g
i
Fig. 3 Basic Network (Open)
b) Check the zero reading of your ammeter with the terminals disconnected, and the zero reading
of your voltmeter with the terminals shorted. If they do not exactly read zero, adjust them with
the small thumbscrew in the center (under the needle indicator) of the meter.
c) Record the values of resistances R1, R2, R3, and R4. Measure and record the values of ε1 and
ε2 . Note: ε1 should be 3 volts and ε2 1.5 volts if the batteries are fresh.

A preceding lab class may have left the batteries in the reversed position
113
Experiment 22
Procedure Part I: Kirchoff's Current Law
d) When the gaps a-b, c-d, and e-f are closed with wires (called jumpers), there are three different
currents in the network: I1in the left loop, I2 in the right loop, and I3 down the center. We assign
them directions given in Fig. 4.
e) To measure I1, use the 500mA scale on your ammeter, connecting it across the gap a-b as
shown. Be careful to connect the + and – terminals of the ammeter properly. This will allow for
either a positive or negative reading.
Close the other two gaps (c-d, and e-f) with jumpers. Your reading determines the true direction
of I1. (See Fig. 4.) If your reading is positive, the current is going from the positive side to the
negative side. Your ammeter dial should read a positive current; if not, make sure that your
batteries are correctly oriented and that your ammeter is correctly connected. If in doubt, call
your instructor.
1
a
A
b
R1
Ammeter
c
d
I1
k
e
I3
f
I2
R3
R2
2
R4
j
g
i
Fig. 4 Basic Network (Closed)
f) If everything is correct, then switch the ammeter to the 50ma scale and record the current I1,
estimating it to 3 significant figures. As soon as you finish this, disconnect the ammeter and
reconnect gap a-b.
g) To measure I2, connect the ammeter across the e-f gap instead of the jumper, observing the
same procedure as in (e) and (f). As soon as you finish this, disconnect the ammeter and
reconnect gap e-f.
114
Experiment 22
h) Repeat all this for I3 using gap c-d. Check your original assumed direction of I3 with the
direction you found for I3 as it appears when read by the ammeter in the gap c-d. As soon as you
finish this, disconnect the ammeter and reconnect gap c-d.
i) According to your measurements, is the equation (1) satisfied or not? If you are not certain,
check with your instructor before proceeding.
Procedure Part II: Kirchoff's Voltage Law
j) On your data-sheet, prepare a table as shown. Set your voltmeter to the 5-volt full scale (Hint:
attach a red connector to the plus (+) terminal of the voltmeter, and a black connector to the –
terminal.) Remember voltage is measured by attaching the voltmeter to the appropriate points
without disrupting the circuit.
k) Read carefully the sign conventions in Section B.
Label resistor + and – in Fig 4 using the true directions
of currents I1 , and I3.
Connect the voltmeter properly as you go around
the loop. Remember that the voltmeter will show you
only the magnitudes of the potential difference. The
signs (plus and minus) must be determined by you.
Measure and record all potential difference magnitudes
and signs in the table. AS SOON AS YOU FINISH
THIS, DISCONNECT ALL GAPS.
Kirchoff's Voltage Law
POTENTIAL
DIFFERENCE
(VOLTS)
SIGN
(+ OR -)
MAGNITUDE
(VOLTS)
Va - Vk
Vk - Vj
Vj - Vc
Vc - Va
ALGEBRAIC
SUM:
l) According to your measurements, is the equation (2) satisfied or not? If you are not certain,
check with your instructor.
Procedure Part III: The Modified Network
m) With all gaps open, connect a wire across the resistor R4. This effectively removes it from the
circuit. Finally, reverse the polarity of emf, ε2 .
115
Experiment 22
n) When the gaps are closed, there will be 3 new values for I1, I2, and I3. (See Fig. 5.) Measure
and record the currents in the Modified Network by the same procedure you used in Part I, to
wit: start with the 500 ma scale as in (d); then proceed as in (e) and (f).
1
a
c
b
A
R1
k
e
d
I'1
I'3
f
I'2
R3
R2
2
Note: R4 is no
longer part of
the circuit.
g
i
Fig. 5 Modified Network (Closed)
BEFORE YOU LEAVE THE LAB:
a) Place ε2 into its original position.
b) Disconnect all jumpers from the circuit board.
c) Make sure you recorded the values of your resistances R1, R2, R3, R4 (they are accurate to
within 0.1%)
d) Make sure you recorded the values of ε1 and ε2, from the voltmeter readings (accurate to
within 1.0%).
Lab Report
Part I: Verification of Kirchoff’s Laws
1) Using your measured values of currents in Basic Network, display the actual value (in mA) of
the left-hand-side of equation (1), when applied to junction g.
2) Using your measured magnitudes and signs of potential differences in Basic Network, display
the actual value of the left-hand-side of equation (2) when applied to loop a-k-j-g-c-a (counterclockwise about the left side loop).
Part II: Basic Network
3) Set up 3 equations with 3 unknowns (the currents) for the Basic Network: one for junction g,
and the other two for loops a-k-j-g-c-a and i-g-c-f-i.
4) Solve these equations by a method of your choice, but show clearly your method and include
most of the steps in your calculation.
116
Experiment 22
WARNING: You are not allowed to use your measured values of currents in any of your
calculations. You may use only the recorded values of resistances and the emf's in order to
determine the theoretical values of the currents.
NOTE: To achieve high grades, solve your equations algebraically. I.e. do not insert any values
for the resistances and emf's. Display a formula for I1, in the following form:
Terms with 's and R's only
I1 =
Terms with R's only
However, once I1 is determined from the algebraic solution, then the remaining currents, I2 and
I3 may be determined by substitution of this value for I1.
5) Display your results in Table 1 as shown. Note that the discrepancies should be "absolute," in
mA, and not in %.
TABLE 1: BASIC CIRCUIT
CURRENT
CALCULATED
(mA)
MEASURED
(mA)
ABSOLUTE
DISCREPANCY
(mA)
I1
I2
I3
Question #1: Why are the internal resistances of the batteries not considered in this experiment?
Part III: Modified Network
6) Set up 3 equations with 3 unknowns, for the modified network, as in (3) above. Display these
modified equations without solving them.
Hint: You should now appreciate the importance of solving the equations algebraically, as in (3),
because by reversing the sign of ε2 and setting R4 = 0 in (3) you automatically have the solution
for the Modified Network.
7) Solve the modified equations. (Note: if you have done (3), just follow the hint above.)
8) Display your results for the modified network in a table similar to Table 1 above.
117