7. Test questions

1. What are the SI units for resistance?

2. What parametres of a conductor define its resistance?

3. How it is possible to define a total resistance in serial and parallel connections of resistors?

4. Draw the scheme of Wheatstone bridge.

5. What is the point of Wheatstone bridge method of measurements of resistance?

6. Derive the relation between resistances of bridge's arms at balance with Ohm’s law.

7. Derive the relation between resistances of bridge's arms at balance with Kirchhoff‘s laws.

8. How it is possible to calculate statistical absolute error and device absolute error of measurements.

8. Content of the report

Homework to Laboratory work №2-2

(Answers on test questions from p.26)

…

Laboratory work № 2-2 implementation protocol

1) Topic: MEASUREMENT OF RESISTANCE WITH WHEATSTONE

BRIDGE

2) Goal: 1. Study the method of measurements by means of a bridge circuit.

2. Study the method of data processing.

3. Finding a resistance of resistor.

3) Scheme of laboratory research facility here Rx – unknown resistor; R1 – etalon resistor; R2 , R3 – resistor boxes, A – zero-indicator, K – switch, e–source of EMF.

4) Table of measuring instruments:

№	Name	Type	Serial number	Grid limit	Grid unit	Accuracy class
1.	Etalon resistor R1	МЛТ	-	10 k	-	β1 = 1 %
2.	One-decade resistors box R2	Р33	-	90 k	10 k	β2 = 0,2 %
3.	Four-decade resistors box R3	Р33	-	9999 k	1 	β3 = 0,2 %
4.	Microaperemeter (zero indicator)

5) Equations for calculation:

5.1) At the ballance condition I_A=0 of Wheatstone bridge the unknown resistance:

.

5.2) Statistical absolute error of measurements:

,

where =0,95 – confidence probability; n=5 – number of measurements;

t _{0,95 ; 5}= 2,77 – Sdudent’s coefficient.

5.3) Device absolute error of measurements:

,

where – average value of measurand; – relative error of measurements; _i – accuracy class of electrical measuring instrument.

5.4) Total absolute error of measurements:

.

5.5) Relative error of measurements

.

6) Table of measurements: R₁ =10 k = 10000 ;

№	R2, 	R3 , 	Rxi ,	Rxi, 	Rxi)2, 2
1.	10 000
2.	30 000
3.	50 000
4.	70 000
5.	90 000
	Average value =		…		…

7) Quantities calculation: …

7.1) Calculation of unknown resistance:

; ; ;

; .

7.2) Calculation of statistical absolute error of measurements of resistance:

.

7.3) Calculation of device absolute error of measurements of resistance:

.

7.4) Calculation of total absolute error and relative error of measurements:

;

8) Final results:

R_x = (<R_x> R_x)_ _; =… %.

9) Conclusions: At measuring of unknown resistance by means of a most accurate method of Wheatstone bridge the basic contribution to an absolute error has introduced … (ststistical or device) error.

10) Data: “___” _____20___. Work done by: ______ Work checked by:

(Surname, readable)

LABORATORY WORK № 2-3

1. Topic: DETERMINATION OF EMF OF CURRENT SOURCE

2. Goal of the work:

2.1. Study of compensating method of measuring the EMF (electromotive force) of modeling sources with a various internal resistance.

2.2. To perform measuring of the EMF of modeling sources with a various internal resistance by a method of direct measuring and to find range of applicability of this method for EMF measuring.

3 Main concepts

In point 2.2 of Main concepts of lab № 2-1 have been given definitions of a potential (11) and potential differences (14) as basic performances of an electric field. Definitions of magnitude of the EMF of a source, a voltage drop and voltage will be below given.

3.1. Potential distribution along the nonuniform circuit unit

Fig. 13 – Potential distribution on terminals of a sourse of the EMF

The nonuniform subcircuit terms the part of an electric circuit containing a source of the EMF. We will consider distribution of a potential on different circuit units with extraneous forces.

At first we will consider a nonuniform electric circuit unit 1-2 consisting only one source of the EMF (see Fig. 13).

In open-circuit (idle) mode the current through a source does not flow I=0, then all work of extraneous force A_EXTR is used only for separation of charges in a source, that is to say for creation of EMF. Thus an open-circuit voltage on the source equals to greatest possible potential difference on source terminals 1-2 (dashed line on fig. 13):

; []=V, (53)

Thus, the EMF of a source  is a greatest possible voltage on its terminals at no-load condition, and numerically is equal to a work done by extraneous forces during a transition of electric charge unit through the source. The EMF of a source is measured in volts.

When through a source  the current I0 flows, then part of work A_EXTR of extraneous forces will be used for overcoming of internal resistance of the source r. As a result, the source heats up. This part of work according to Joule-Lentz law is equal:

A_RES=I²rt=qIr.

Then voltage on the source terminals 1-2 will be less, than in idle mode conditions (53) on value of voltage drop Ir on internal resistance of the source:

. (54)

On a Fig. 13 this voltage drop Ir on internal resistance of the source r is denoted by an arrow downwards.

Fig. 14 – Potential distribution along a nonuniform circuit unit

Let's consider a nonuniform circuit unit 1-3, which contains the source  and an external (load) resistor R (see Fig. 14). Then part of work A_EXTR of extraneous forces will be used for overcoming of internal resistance of the source r and resistance of load resistor R. As a result, the source and the external resistor heat up. Therefore, this part of work we can obtain using a Joule-Lentz law:

A_RES=I²(r+R)t=qI(r+R).

Then voltage on circuit unit terminals 1-3 in comparison with 1-2 (54) is being decreased by value of voltage drop Ir, and, in addition, by value of voltage drop IR on external resistor R:

. (55)

On a Fig. 14 both voltage drops Ir on internal resistance of the source r and on load resistor R are denoted by arrows downwards.

Fig. 15 – Potential distribution along a nonuniform circuit unit, which contain opposite connection of EMF

Let's consider a circuit unit 1-4, which contains the source ₁, which action of an extraneous forces coincides with a current direction, external resistance R, and a source ₂, which action of an extraneous forces does not coincide with a current direction (see Fig. 15).

In case ₂<₁, it means, that the source ₂ will be charged by a current, created by the source ₁. Then part of work A_EXTR of extraneous forces will be used for overcoming of internal resistance of the source r₁, for overcoming of resistance of load resistor R, for overcoming of internal resistance of the second source r₂:

A_RES=I²(r₁+R+r₂)t=qI(r₁+R+r₂),

and, in addition, for source ₂ charging:

A_CH=₂ q.

Therefore voltage on circuit unit terminals 1-4 in comparison with 1-3 (55) is being decreased by value of voltage drop Ir₂ on internal resistance of second source, and in addition by value of it EMF ₂:

. (56)

As we can see, ₁ is positive, because direction of extraneous forces work (from “–“ to “+” inside the source) coincides with direction of current I, but ₂ is negative, because direction of extraneous forces work noncoincides with direction of current I.

Рис. 16 – Potential distribution along a closed circuit

If terminals 1 and 3 of nonuniform circuit unit 1-3 that is shown on Fig, 14 are closed by a conductor, then potentials of points 1 and 3 will be equalized:

₁=₃,

and we obtain the elementary variant of closed electric circuit (see Fig. 16). In this case all work A_EXTR=q of extraneous forces will be used for overcoming of internal resistance of the source r and resistance of load resistor R:

A_EXTR= A_RES= qI(r+R).

From (55) we obtain, then all EMF of the source  drops on internal resistance of the source r and on external resistor R:

.(57)

From formulas (53) - (56) we can see, that according to definition the voltage U – is a potential difference on terminals of a circuit unit.

However, it is necessary to note, that voltage U_r on terminals of an EMF source (54) is being accepted with plus sign, and is being defined as work of extraneous (internal) forces on transition of a unit positive charge. Thus, from a Fig.14 we can see, that it is equal U_r=₂–₁ to a difference of a terminating potential ₂ (greater), and an initial potential ₁ (smaller). Voltage on any nonuniform circuit unit (53) - (56) is being analogously defined.

If we define a voltage U_R on terminals of the homogeneous circuit unit 2-3, which is not containing an EMF source (Fig. 14), it will be positive, when we consider it as a work of field forces (external) on transition of a unit positive charge. Thus, from a Fig. 14 we can see, that it is equal U_R=₂–₃ to a difference of an initial potential (greater) ₂, and a terminating potential (smaller) ₃:

. (58)

Therefore for the homogeneous circuit unit AB from fig. 10 Ohm's law write down in the form of the formula (20).