Microsoft Word - Lab 6_316_2021_1.docx
Lab Section # 6
Kirchhoff’s Law:
Voltage and Cu
ent
In
Circuits
2
6-Kirchhoff’s Law: Voltage and Cu
ent in Circuits
Purpose:
To become familiar with the operation of the Wheatstone
idge. To use the Wheatstone
idge to
test the laws of resistance of metal wire.
Apparatus:
Wheatstone
idge board, galvanometer, resistance box XXXXXXXXXXΩ range, several spools of
esistance wire of known diameter and known length, power supply, 10 Ω rheostat, hook up wire.
Theory:
One of the difficulties encountered in determining the resistance of a circuit from
measurements made with voltmeters and ammeters is that these instruments do affect the circuit.
Whenever high precision is required, the effect of the meters in the circuit must be taken into
account. The Wheatstone
idge circuit, designed about 1833, enables the determination of the
esistance of a circuit element without distu
ing the circuit. The circuit diagram for the
Wheatstone
idge is shown on the right.
Let X (also known as Rx) be an unknown resistance and let R1, R2, R3 be resistors whose values
are known. To determine the value of Rx, one chooses R1, R2, R3 so that the galvanometer reads
zero. When this happens, the cu
ent through R1 and Rx is the same as the cu
ent through R2 and
R3, and we say the
idge is ‘balanced”.
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When the
idge is balanced, the potential difference across the galvanometer is zero. Let’s look
at what happens when the
idge is balanced. This means that the potential drop across resistance
R1 must be equal to the potential drop from across resistance R2. If we designate the cu
ent in the
upper
anch as I1, and the cu
ent in the lower
anch as I2, we would get from applying Ohm's
law to resistors R1 and R2 that
I1R1=I2R2.
The fact that the potential drop across the
idge is the same whether we consider the upper
anch
or lower
anch together with the information from the above equation allows us to write a second
equation.
I1Rx=I2R3.
We now divide the second equation by the first we obtain.
and solving for Rx gives us
The important feature about this result is that it makes possible the determination of Rx without
knowing any cu
ent or the voltage across any resistors. It is true we needed a galvanometer but
we used it to tell us when there no cu
ent through it so its presence in the circuit did not alter
conditions in either
anch of the circuit. As this equation indicates, one needs it to know only
the ratio R3/R2 and the value of R1 in order to determine Rx.
The set up you will use is shown on the right. R2 in this diagram is represented by a length of wire
L1. R3 is represented by the length of wire L2. We will assume that the wire has uniform density
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and uniform cross section, so
The experiment is ca
ied out by pressing the
high sensitivity button on the galvanometer (not
shown, but you used it last week) and
moving contact point B along the wire until the
galvanometer shows zero cu
ent. Again, when
this condition is achieved, the Wheatstone
idge is said to be "balanced".
The law of resistance for metallic
conductors states that the resistance of a metal
wire is given by the equation (you used this
last week also)
where L is the length of the wire, A is the cross sectional area of the wire. ρ (rho) is called the
esistivity of the wire and is determined by the material of which the wire is made. To test the law
of resistance one could make some measurements and check if the above equation is valid. In this
experiment spools of wires will be used, you will be given L, and asked to measure A and R. Then
you will check your value for ρ against the given value.
In some cases, the manufacturer indicates that these wires are alloys of nickel and silver and we
may not know the percentage composition or the value of the resistivity, ρ. If we have two spools
of wire made of the same material then we could write.
divide these two equations gives
where dβ and dα are the respective diameter of the two wires. This is an example of how you
can test the law of resistance.
Procedure:
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A diagram of the equipment is shown at the right. R1 is resistance box. R0 is a 10Ω or so rheostat
that maybe needed in order to be able to limit the cu
ent in the circuit. L1 and L2 is one piece of
wire mounted on a meter stick, which in turn, is mounted on a board. Set the power supply at
about .2 amps.
Rx will be a spool of wire. Most of boards used for this experiment are made up of an alloy of
nickel and silver, one or more spools may be copper. We suggest you select three different spools.
There are a limited number of these so the various partner groups may need to share this spool.
After the circuit has been connected, set R1 to some value between 0 Ω and 10 Ω, press the low
sensitivity button on the galvanometer and move B until the galvanometer shows zero deflection.
Then press the middle sensitivity button, readjust B to zero the galvanometer, and finally the most
sensitive button. Move B again to zero the galvanometer. This procedure, using all three buttons,
helps protect the meter. If the needle does not move wildly, then you can start with the medium
or sensitive scale. How sensitive the circuit is will depend upon the amount of cu
ent which is
controlled by the setting of Ro. Change the vale of R1 and repeat the procedure.
Strong Hint: Make your new value of R1 close to your value of Rx, that way, the
idge will be
alanced near the middle. Make three or four determinations for the resistance of each coil of wire.
This will enable you to determine an average value for the resistance as well as the precision of
each value. When your values of R1 are less than 10 Ω, double check them with a D.V.M.
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Wheatstone
idge data sheet
NAME ________________________ Lab Time ___________
SPOOL #1 (Cu)
Material
Copper
Diameter (m)
Area (m2)
Gauge
----->
Length (meters)
R1 (plug box)
(ohms)
L1
(meters)
L2
(meters)
Rx
(ohms)
Reading 1
10
--->
Reading 2
9
--->
Reading 3
8
--->
Reading 4
7
--->
Rx= _____________ ρ=___________ ± _______% Δ% from 2.18x10-8 Ωm ______
8
SPOOL #2 (Cu)
Material
Copper
Diameter (m)
Area (m2)
Gauge
----->
Length (meters)
R1 (plug box)
(ohms)
L1
(meters)
L2
(meters)
Rx
(ohms)
Reading 1
10
--->
Reading 2
9
--->
Reading 3
8
--->
Reading 4
7
--->
Rx= ______________ ρ=___________ ± ______% Δ% from 2.18x10-8 Ωm=_________