HOW DO WE MEASURE CURRENT,

VOLTAGE AND RESISTANCE?

DEVICES USED TO MEASURE:

CURRENT-AMMETER VOLTAGE-VOLTMETER RESISTANCE-OHMMETER

VOLTMETER OHMMETERAMMETER

CURRENT VOLTAGE RESISTANCE

Most of the time these three meters are combined to form a MULTIMETER that can be switched from one to another. There are 2 types of such meters: those who use digital electronics and those that do not.

FOR THE NON-DIGITAL DEVICES:

GALVANOMETER A galvanometer is a type of

sensitive ammeter: an instrument for detecting electric current. It is an analog electromechanical transducer that produces a rotary deflection of some type of pointer in response to electric current flowing through its coil in a magnetic field. It was named after the italian anatomist Luigi Galvani who became famous during his time because of his experiments on “animal electricity”using frogs as specimen.

AMMETER

The ammeter is a galvanometer with a low resistance, called shunt resistor, connected parallel to it.

The shunt resistor provides a bypass for the current in excess of the galvanometer’s full scale limit.

Sample problem 11.12

A galvanometer has a coil resistance of 50.0Ω and fullscale current of 10mA. Convert this to an ammeter reading 1.0 A full scale.

GIVEN: resistance= 50 Ω Current= 1.0 mA Scale= 10 ARECQUIRED: Convert this to an ammeterSOLUTION:To convert the galvanometer to an ammeter, we have we

have to connect a resistor Rp parallel to it. Let Ip and Vp be the current passing through the voltage across resistor Rp, respectively.

SOLUTION:To convert the galvanometer to an

ammeter, we have we have to connect a resistor Rp parallel to it. Let Ip and Vp be the current passing through the voltage across resistor Rp, respectively. Let RG , IG and VG represent the resistance, the current passing through and the voltage across the galvanometer, respectively. We are given that RG =50.0 Ω and

IG =1.0mA= 0.001 A. We are also given that the total current I= 10.0A.

by law of parallel scissors,

by law of parallel scissors, I= IG + IP

10 A = 0.001 A+ IP

solving for IP,

IP = 9.999A

again by law of parallel scissors,VT = VP = VG

using ohm’s law,IP RP = IG RG

substituting values,

(9.999A) RP = ( 0.001 A) (50 Ω)

RP = 0.005 Ω

Sample problem 11.13

Convert the galvanometer in the preceeding example to a voltmeter reading 3.0 V full scale.

GIVEN: Resistance= 50 Ω Current= 0.001 ARECQUIRED: Convert to galvanometer

SOLUTION: Using ohm’s law we determine the maximum

voltage it can measure.V=IRV=(0.001 A)(50 Ω)V=0.05V To convert this to a voltmeter that can read up

to 3.0 V, a resistance Rs must be connected in series with the galvanometer,

IG = IS = 0.001 A

Vtotal = VG + VS

Vs = 2.95 V

But VS = IS RS

2.95 V= (0.001 A) RS

solving for RS,

Rs =2950 Ω

Resistance may be measured directly by an ohmmeter. A simple ohmmeter consists of a galvanometer in series with a battery and a resistor R1 , as shown in the fig 11-10. The value of R1 is chosen so that when the 2 terminals x and y are made to touch each other, the galvanometer gives full-scale deflection. A full scale deflection of the galvanometer indicates zero resistance between x and y. when x and y are connected across an unknown resistance R, the galvanometer will deflect depending on the value of R. hence the scale can be calibrated to measure the resistance.

The voltmeter-ammeter method of determining resistance requires that the current passing through the resistor as well as the potential difference across the resistor be measured by an ammeter and a voltmeter, respectively. The resistance is computed using ohm’s law.

The Wheatstone bridge is an electrical circuit used for comparing resistances. It was invented by Hunter Christie but named after sir Charles Wheatstone who popularized the use of the electrical circuit. It consists of a battery, galvanometer and four resistors. The 4 resistors are divided into 2 parallel branches, each brand consisting of 2 series resistors. Of the four resistors, 3 are of known while one is unknown. To determine the value of the unknown resistor , the known resistances are adjusted and balanced until the current passing through the galvanometer is zero. The unknown resistance may be found using the following relation.

SAMPLE PROBLEM 11.14

An unknown resistance RS is being measured by means of a Wheatstone bridge. The galvanometer reads zero when the other resistors R1, R2 and R3 are 3 Ω, 2 Ω and 10 Ω, respectively. Find the unkown resistance Rx . Refer to the figure 11-11.

GIVEN: R1= 3 Ω, R2 = 2 Ω ,R3 = 10 V

REQUIRED: Rx

SOLUTION:Rx = R1

R3 R2

Rx = 3.0 Ω

10.0 Ω 2 ΩRx = 15 Ω

TAPOS!!!