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Current Electricity

Unit II: Current Electricity

Dr Sukanta Deb

Department of Physics, Cotton UniversityPanbazar, Guwahati (Assam)

Subject: Physics, HS 2nd Year

Module: IV

Dr Sukanta Deb Unit II: Current Electricity 1 / 51

Current Electricity

Outline

1 Current Electricity


Current Electricity

Learning Objectives (Module: IV)

1 GalvanometerConversion of a Galvanometer to an AmmeterConversion of a Galvanometer to a Voltmeter

2 PotentiometerPrinciple and Applications to Measure PotentialDifferenceComparing emfs of Two CellsMeasurement of Internal Resistance of a Cell


Current Electricity

GalvanometerGalvanometer is an instrument which is usedto detect the flow of electric current in acircuit and its direction of flow. The crucialpart of an analog ammeter or voltmeter is agalvanometer. A galvanometer gives readingby a pointer on a scale.

Galvanometer consists of a coil of wire inthe magnetic field of a permanentmagnet.When there is a current in the coil, themagnetic field exerts a torque on the coil,which causes the coil to rotate.A pointer attached to the coil indicatesthe reading on a scale. The coil itselfcontributes some resistance when thegalvanometer is placed within a circuit.

When no currentpasses through it, theneedle stays in themiddle of thegraduated scale. Thispoint is marked zero.


Current Electricity

GalvanometerCurrent can be passed through a galvanometer in eitherdirection. The needle deflects accordingly towards left ortowards right depending on the direction of the current flow.The deflection of the needle of a galvanometer is directlyproportional to the current flowing through it, i.e., I ∝ φ.The full-scale current sensitivity of a galvanometer, is theelectric current needed to make the needle deflect full scale.This is also called full scale deflection current of thegalvanometer. A galvanometer whose sensistivity is 50 µAcan measure currents from 0 µA up to 50 µA.A galvanometer is useful only for the measurement of small dccurrents and dc voltages.

Symbol of a galvanometer:


Current Electricity


Current Electricity

AmmeterThe device that measures current in an electric circuit is called anammeter. Some of the ammeters with various current measuringcapacities used in the laboratory are shown in the following figures.

For measuring current in an electric circuit, the ammeter has to beconnected in series. Otherwise, the actual current to be measuredwill get divided if placed in parallel. An ideal ammeter should havezero resistance. Current can be passed through an ammeter only inone direction from higher potential to lower potential. When nocurrent passes through an ammeter, the neddle stays at zero whichis marked at the extreme left of the scale.


Current Electricity

VoltmeterThe device that measures potential difference (voltage drop)between two points in an electric circuit is called a voltmeter. Avoltmeter capable of measuring potential difference with variousranges is shown in the following figure.

A voltmeter measures only a positivevalue of the potential difference (pd).Higher potential has to be connectedto the positive terminal and the lowerpotential to the negative terminal ofthe voltmeter.For measuring pd between two points,the voltmeter has to be connected inparallel. Otherwise, the actual pd tobe measured will be divided if placedin series.

An ideal voltmeter shouldhave an infinite resistance.When the pd is zero, theneddle stays at zero which ismarked at the extreme leftof the scale.


Current Electricity

A galvanometer alone can’t be used to measure current?

Let us consider a circuit containing abattery of emf E and a resistance R. Thecurrent I flowing in the circuit given by

I = ER .

In order to measure this current we connecta galvanometer in series with R. Let r bethe resistance of the galvanometer. Thecurrent flowing in the circuit now becomes

I ′ = ER + r < I .

Therefore the galvanometer gives a value ofthe current less than the required current I

that we wanted to measureand hence results into anerror. So a galvanometercannot be used directly tomeasure current.


Current Electricity

In order to reduce the error, we should make the resistance ofthe galvanometer very small such that r << R and hence rcan be neglected, then we will have I ′ ≈ I .The idea of conversion of a galvanometer to an ammeterarises from this fact.In order to use a galvanometer as an ammeter, we shoulddecrease its resistance by introducing a low resistance called‘shunt’ in parallel with it.In this case the effective resistance of the galvanometerdecreases which gives I ′ ≈ I .


Current Electricity

Pictorial Representation: Galvanometer to an Ammeter

Figure: A galvanometer with afull scale deflection currentIG = 1 mA. The galvanometerwas able to measure currentfrom 0 to IG, where IG = 1 mAand corresponds to full scaledeflection of the galvanometer.

Figure: A galvanometer with a fullscale deflection current I = 1 A.When a low value resistance Rshcalled shunt (chosen after duecalculation) is placed in parallel withthe galvanometer, the galvanometercan measure large value of currentfrom 0 to I , where I >> IG. Here Icorresponds to full scale deflectionof the galvanometer. In the presentcase I = 1 A.


Current Electricity

Construction of an AmmeterLarger currents can be measured with an ammeter. An ammeterconsists of a galvanometer in parallel with a resistor called shuntresistor (“shunt” is a synonym for “in parallel” ) as shown in thefollowing figure.

The shunt resistance is Rsh and the resistance of the galvanometercoil, through which current passes, is r . Usually Rsh << r suchthat the equivalent resistance of the ammeter Req = rRsh

r+Rsh< Rsh

is small so that the majority of the current passes through theshunt resistor and very little current passes through the galvano-meter to deflect the needle.


Current Electricity

Construction of an Ammeter

We know that a galvanometer can measure very small dc current.A galvanometer with a full scale sensitivity of IG can measurecurrents only up to IG , where IG is very small. Here IG denotesthe current corresponding to the full scale deflection of thegalvanometer. But we want the galvanometer to measure currentup to I , where I >> IG for the full scale deflection. Let I

IG= n,

where n is called the range. In order to measure current up to I ,we place a very small resistance Rsh called the shunt resistance inparallel with the galvanometer. The galvanometer is now called anammeter. The maximum value of the current I that can bemeasured for full scale deflection of the galvanometer is passedthrough the ammeter by placing it in series. From the circuit wehave

I =IG + IR.


Current Electricity

Since r and Rsh are parallel to each other, both will have the samepotential difference,

IGr =IRRsh

⇒ Rsh =IGIR

r

⇒ Rsh =( IG

I − IG

)r

⇒ shunt resistance =( IG

I − IG

)× resistance of the galvanometer

This is the required value of the shunt resistance in order toconvert a galvanometer to an ammeter. Also we can write

rRsh

= IRIG

⇒ r + RshRsh

=IR + IGIG

(∵

ab = c

d ⇒a + b

b = c + dd

)⇒ r + Rsh

Rsh= I

IGDr Sukanta Deb Unit II: Current Electricity 15 / 51

Current Electricity

⇒ r + RshRsh

=n

⇒ r + Rsh =nRsh ⇒ r = (n − 1) Rsh

⇒ Rsh = rn − 1

⇒ shunt resistance =galvanometer resistancerange− 1 .

The equivalent resistance of the ammeter is

Req =r ||Rsh = rRshr + Rsh

⇒ Req = rr+Rsh

Rsh

⇒ Req = rn

⇒ equivalent resistance =galvanometer resistancerange .


Current Electricity

1 A galvanometer has a resistance of 50 Ω and its full scaledeflection current is 50 µ A. What value of shunt is to beused so that it can measure currents upto 5 mA?Given r = 50 Ω, IG = 50 µ A, I = 5 mA, Rsh =? We knowthat range is given by

n = IIG

= 5 mA50 µ A

⇒ n = 5× 10−3

50× 10−6 = 110 × 10−3 × 106 = 100.

The shunt resistance is given by

Rsh = rn − 1 = 50

100− 1 = 0.5 Ω.


Current Electricity

2 An ammeter is to be constructed which can read currents upto 2.0 A. If the coil has a resistance of 25 Ω and takes 1mAfor full-scale deflection, what should be the resistance of theshunt used?Given I = 2.0 A, r = 25 Ω, IG = 1 mA, Rsh =?We know that the range (n) is given by

n = IIG

= 2.0 A1 mA

⇒ n = 2.0 A10−3 A = 2× 103 = 2000.

The shunt resistance Rsh is given by

Rsh = rn − 1

⇒ Rsh = 252000− 1

⇒ Rsh =12.5× 10−3 Ω⇒ Rsh =1.25× 10−2 Ω.


Current Electricity

3 The deflection in a moving coil galvanometer falls from 50divisions to 10 divisions when a shunt of 12 Ω is applied.What is the resistance of the galvanometer, assuming maincurrent is the same?We know that the current in a galvanometer is directlyproportional to deflection of the galvanometer, i.e., I ∝ φ.Let I be the main current. Then the galvanometer currentIG = I

5 . Hence

IIG

=5⇒ n = 5.

The resistance of the galvanometer

r = (n − 1) Rsh

⇒ r = (4× 12) Ω⇒ r =48 Ω.


Current Electricity

A galvanometer alone can’t be used to measure voltage?Let us consider a circuit containing abattery of emf E and a resistance R.The battery sends a current I in thecircuit given by

I = ER .

The potential difference across theresistance R is given Ohm’s law

V =IR.

In order to measure this voltage weconnect a galvanometer in parallel withR. The current now gets divided at thejunction into two parts:

IG through the galvanometerand (I − IG) through R:

I =IG + (I − IG) .


Current Electricity

Let r be the resistance of the galvanometer. The voltage read bythe galvanometer is

V ′ =IGr = (I − IG)R⇒ V ′ =IR − IGR = V − IGR⇒ V ′ <V .

Therefore the galvanometer gives a value of the voltage less thanthe required voltage V that we wanted to measure and henceresults into an error. Hence a galvanometer alone cannot be useddirectly to measure votage or potential difference. In order toreduce the error, we should the make the branch current IG verysmall such that IGR << IR and hence IGR can be neglected,then we will have V ′ ≈ V . The idea of conversion of agalvanometer to a voltmeter arises from this fact. In order to use agalvanometer as a voltmeter we should increase its resistance byintroducing a high resistance value in series with it. In this case,the current IG flowing though the galvanometer will be negligibleand hence (I − IG) ≈ I . This will give V ′ ≈ V .


Current Electricity

Pictorial Representation: Galvanometer to a Voltmeter

Figure: A galvanometer with an internalresistance of r = 50 Ω and a full scaledeflection current IG = 1 mA canmeasure voltage from 0 to V0, where

V0 =IGr=(1 mA× 50 Ω)=50 mV

Figure: A galvanometer with an internalresistance of r = 50 Ω and a full scaledeflection current IG = 1 mAconnected with a series resistance ofhigh value Rser = 5 kΩ can measurevoltage from 0 to V , where

V =IG (r + Rser)= [1 mA× (50 + 5000) Ω]=5.05 V.


Current Electricity

Construction of a Voltmeter

A voltmeter consists of a galvanometer in series with a resistorhaving a very high resistance value so that the equivalentresistance of the voltmeter is much larger than the resistance ofthe galvanometer coil alone. The construction of the voltmeter isshown in the following.

Let V0 = IGr be the maximum voltage that can be measured bythe galvanometer for the full deflection current IG of thegalvanometer. But we want the galvanometer to measure voltageup to V , where V >> V0 for the full scale deflection. Let V

V0= n,

where n is called the range.


Current Electricity

In order to measure voltage up to V , we place a very highresistance Rser in series with the galvanometer. The galvanometeris now called a voltmeter. The maximum value of the voltage Vbetween two points that can be measured for full scale deflectionof the galvanometer is passed through the voltmeter by placing itin parallel. From Ohm’s law

⇒ V =IG (Rser + r) .

ThereforeVV0

=Rser + rr ⇒ n = Rser + r

r⇒ nr =Rser + r ⇒ Rser = (n − 1) r

⇒ series resistance = (range− 1)× galvanometer resistance.

The equivalent resistance of the voltmeter is

Req =Rser + r = (n − 1) r + r = nr⇒ equivalent resistance =range× galvanometer resistance.


Current Electricity

1 A voltmeter consists of a 25 Ω coil connected in series with a575 Ω resistor. The coil takes 10 mA for full scale deflection.What maximum potential difference can be measured on thisvoltmeter?Given r = 25 Ω, Rser = 575 Ω, IG = 10 mA = 10−2 A. Themaximum potential difference that can be measured on thisvoltmeter is

V =IG (Rser + r) = 10−2 (575 + 25) = 10−2 × 600 = 6 V.


Current Electricity

2 A voltmeter coil has resistance 50 Ω and a resistor of 1.15 kΩconnected in series. It can read potential difference up to12 V. If this same coil is used to construct an ammeter whichcan measure currents up to 2 A. What should be theresistance of the shunt used?For the voltmeter:r = 50 Ω, Rser = 1.15 kΩ = 1.15× 103 Ω = 1150 Ω,V = 12 V. Let IG be the full scale deflection current of thegalvanometer. We know that

V =IG (Rser + r)

⇒ IG = VRser + r = 12

1150 + 50

⇒ IG = 121200 = 10−2 A.

If the same coil is used for the construction of an ammeter,we have r = 50 Ω, IG = 10−2 A, I = 2 A. Thereforen = I

IG= 2

10−2 = 200 A.Dr Sukanta Deb Unit II: Current Electricity 26 / 51

Current Electricity

We know that the shunt resistance is given by

Rsh = rn − 1 = 50

200− 1 = 0.251 Ω.


Current Electricity

PotentiometerPotentiometer is a device which doesnot draw any current from the givencircuit and still measures the potentialdifference. Thus, it is equivalent to anideal voltmeter. It is mainly used tomeasure emf of a given cell and tocompare emf’s of cells. It is also usedto measure internal resistance of a givencell.

A potentiometer consists of a longuniform wire AB usually 5 to 10meters long made of constantan ormanganin connected in series bythick copper strips stretched on awooden board. The ends terminalsA and B of the wire are connected

to a battery of known emf Eswith an ammeter, a key and arheostat (variable resistance) inseries. When the key K is closed,the battery sends the current Ithough the potentiometer wireAB which is kept constant byusing the rheostat. This is calledthe auxiliary/primary circuit.


Current Electricity

PotentiometerWe assume that the standard battery (also called driving cell) haszero internal resistance. Let L,A, ρ and R denote the length, areaof cross-section, resistivity of the material and resistance, of thewire AB, respectively. The current flowing through thepotentiometer wire is given by

I = EsRh + R .

The voltage (potential drop) across the potentiometer wire oflength L is given by

V (L) =IR

Let k be the potential drop per unit length (potential gradient) ofthe wire. k is given by

k =V (L)L [V/cm].


Current Electricity

k can also be written as

k =V (L)L = IR

L = R IL

⇒ k =(ρLA

)IL

⇒ k =ρIA .

If I and A are constants then k is also a constant.Therefore the potential drop across a segment of length x is givenby

V (x) = kx⇒ V (x) ∝ x.

This forms the principle of a potentiometer.


Current Electricity

PotentiometerThus the principle of a potentiometer can be stated as “thepotential drop across any segment of length of the wire isproportional to the length of that segment provided the wire is ofuniform area of cross-section and a constant current is flowingthough it”.Now we will make the secondary circuit. In order to do that we willbring the cell whose emf E is to be determined. The positiveterminal of this cell is connected to the end A of the potentiometerwire. The negative terminal of the cell is connected in series to agalvanometer and then to a jockey. The jockey may be slid on thewire AB and may touch the wire at any desired point. The lowercircuit shown in the earlier figure forms the secondary circuit. Letthe jockey touch the wire at any point J which is at a distance xfrom the end A of the wire. Therefore the potential drop acrossthe segment of length AJ = x of the wire is

V (x) =kx.Dr Sukanta Deb Unit II: Current Electricity 31 / 51

Current Electricity

PotentiometerThe net emf of the secondary circuit is given by (applyingKirchoff’s loop law)

Net emf =E − kx

Now let us consider the case when x = 0 (i.e the jockey is broughtto A). Then kx = 0. In this case net emf=E .


Current Electricity

Since the circuit is closed, current will flow through the secondarycircuit clockwise from the positive terminal of the cell to thenegative terminal and the galvanometer will deflect towards theleft.When the jockey is moved towards B slowly, x will increase slowlyand hence kx. As a result net emf will decrease. The deflectionwill still be towards left but with some decreased value. On furtherincreasing x, a stage comes when E and kx become equal and thenet emf becomes zero. In this case, the galvanometer will shownull deflection. Let this value of x be equal to l which can bemeasured from the meter scale. l is called the balancing length.Therefore at x = l, we have

E − kl =0⇒ E =kl.

Since l and k are known, the emf E of the unknown cell can bedetermined using this formula.


Current Electricity

Comparison of emfs of Two Cells

A battery of known emf Es is connectedbetween the two end terminals of thepotentiometer wire with an ammeter, akey and a rheostat in series. When thekey K is closed, the battery sends thecurrent I though the potentiometerwire AB which is kept constant byusing the rheostat. This circuit is calledthe primary/auxiliary circuit. Two cellsof emfs E1 and E2 to be compared areconnected to the primary circuit asshown in the figure. The positiveterminals of both the cells are connectedto the end A of the potentiometer wire.

Their negative terminals areconnected to a galvanometer anda jockey via two separate keys K1and K2, respectively. First thekey K1 is closed (i.e., connected)and K2 open (i.e not connected)to include the cell of emf E1


Current Electricity

into the circuit. The jockey is moved over the wire until thegalvanometer gives a null deflection. Let this position of the jockeybe J1. Then the balancing length will be AJ1 = l1. Therefore wehave

E1 =kl1,

where k is the potential gradient of the wire. Now, the key K2 isclosed and K1 left open to put the cell of emf E2 into the circuit.The jockey is again moved over the wire until the galvanometershows a null deflection. Let this position of the jockey be J2. Thebalancing length will be AJ2 = l2. Therefore we have

E2 =kl2.

ThereforeE1E2

= l1l2.

Thus the ratio of the two balancing lengths gives the ratio of thetwo emfs.


Current Electricity

Measurement of Internal Resistance of a CellA battery of known emf Es is connectedbetween the two end terminals of thepotentiometer wire with an ammeter, akey and a rheostat in series. When thekey K is closed, the battery sends thecurrent I though the potentiometerwire AB which is kept constant byusing the rheostat. This circuit is calledthe primary circuit. The cell of emf Ewith internal resistance r to bemeasured is connected to the primarycircuit as shown in the figure. Thepositive terminal of the cell is connectedto the end A of the wire. The negativeterminal of the cell is connected to ajockey through a galvanometer in series.

A known resistance R isconnected across the cellthrough a key K1. With K1open (i.e., not connected),the jockey is moved over thepotentiometer wire until thegalvanometer shows nulldeflection. Let this position ofthe jockey be J1. Thebalancing length AJ1 = l1 ismeasured.


Current Electricity

Since the cell is in open circuit (I = 0), its emf is given by

E =kl1.

Now the key K1 is closed so that the resistance R is introduced inthe cell circuit and the cell is now in a closed circuit. The cellsends a current I1 through the resistance R. Since I1 , 0, what wecan measure is the terminal potential difference across the cell andnot its emf. The jockey is now moved over the potentiometer wireuntil the galvanometer shows no deflection. Let the position of thejockey at this position be J2. The balancing length AJ2 = l2 ismeasured. Then the terminal potential difference of the cell isgiven by

V =kl2.

This is equal to the voltage across R,i.e., V = I1R.


Current Electricity

Applying Kirchoff’s loop law to the right hand panel of the figure,we have

E − I1R − I1r =0⇒ E −V − I1r = 0

⇒ I1r =E −V ⇒ r = E −VI1

= E −VVR

⇒ r =(E −V

V

)R =

( EV − 1

)R

Therefore we have

r =(kl1

kl2− 1

)R

⇒ r =( l1

l2− 1

)R.

Thus knowing the values of l1, l2 and R, the internal resistance rof the cell can be determined.


Current Electricity

1 A uniform wire of length 400 cmand resistance 2Ω is used as apotentiometer. The wire isconnected in series with a batteryof emf 5 V and an externalresistance of 3 Ω. With a cell ofunknown emf E , the balancinglength is found to be 250 cm. Find(a) the potential gradient along thepotentiometer wire and (b) thevalue of E .Given L = 400 cm, R = 2 Ω,Es = 5 V, Rh = 3 Ω, l = 250 cm.The current flowing through thepotentiometer wire is given by

I = EsRh + R = 5

3 + 2 = 1 A.

(a) Potential drop across AB :

V =IR = (1 A) (2 Ω) = 2 V.

Potential gradient along AB

k =VL = 2 V

400 cm⇒ k =0.005 V/cm

(b) ∴ E =kl = 0.005× 250 = 12.5 V.


Current Electricity

2 In a potentiometer arrangement a cell of emf 1.24 V gives abalance point at 35.0 cm legth of the wire. If the cell isreplaced by another cell and the balance point shifts to63.0 cm, what is the emf of the second cell?Here E1 = 1.24 V, l1 = 35.0 cm and l2 = 63.0, then E2 =?We know that

E2E1

= l2l1

⇒ E2 = l2l1E1

⇒ E2 =63.035.0 × 1.24

⇒ E2 =2.25 V.


Current Electricity

3 In the potentiometer circuit as shown,find the value of l when thegalvanometer shows no deflection.The length of the wire is 100 cm andits resistance is 3 Ω.Given L = 100 cm, Rh = 12 Ω,Es = 3 V, r = 1.5 Ω, R = 3Ω. Thecurrent flowing through thepotentiometer wire is given by

I = EsRh + R = 3

12 + 3 = 0.2 A.

Potential difference across AB = L is

V (L) =IR = 0.2× 3 = 0.6 V.

Therefore potential gradient

k =V (L)L = 0.6

100 = 0.006 V/cm.

Potential difference across ldue to Es

V (l) =kl = 0.006l.


Current Electricity

At the balancing length the galvanometer shows null deflection andhence there is no current flowing through the primary circuit of thepotentiometer. However in the secondary circuit, the cell with emfE = 2 V sends a current I1 (say) through the resistance 1.5 Ω and0.5 Ω as the circuit is a closed circuit. The current flowing in thiscircuit is given by

I1 = E1.5 + 0.5 = 2

2 = 1 A.

Therefore the potential difference across 0.5 Ω is

=1× 0.5 = 0.5 V.

At l, the potential difference across l due to E is

V ′(l) =0.5 V.

At balance point

V (l) =V ′(l)⇒ 0.006l =0.5⇒ l = 83.3 cm.


Current Electricity

4 The length of a potentiometer wireis 600 cm and it carries a currentof 40 mA. For a cell of emf 2 Vand internal resistance 10 Ω, thebalancing length is found to be500 cm. If a voltmeter is connectedacross the cell, the balancinglength is decreased by 10 cm. Findthe resistance of the wire, (b) theresistance of the voltmeter and (c)the reading of the voltmeter..Given AB = L = 600 cm,I = 40 mA = 0.04 A,AJ1 = l1 = 500 cm, andAJ2 = l2 = 490 cm. Let R be theresistance of the potentiometerwire AB of length L.

The potential difference acrossAB is

V (L) =IR = 0.04R.

Therefore the potentialgradient along AB is

k =V (L)L

⇒ k =0.04R600 V/cm.


Current Electricity

Therefore the potential difference across AJ1 = l1 due to thedriving cell of emf Es is

Vl1 =kl1 = 0.04R600 × 500

⇒ Vl1 =0.04R6 × 5.

When the voltmeter is not connected, the potential differenceacross AJ1 due to cell of emf E is equal to 2 V since no current isdrawn from the cell at the balance point. Hence

Vl1 =2⇒ 0.04R6 × 5 = 2⇒ R = 60 Ω.

Let RV be the resistance of the voltmeter. When it is connectedacross the cell of emf E = 2, the current drawn from it is given by

I1 = 2RV + 10 .


Current Electricity

Potential difference across the voltmeter is

VRV =I1RV = 2RVRV + 10 .

This must be equal to the potential difference across AJ2 due tothe driving cell of emf Es which is given by

Vl2 =kl2 = 0.04R600 × 490

⇒ Vl2 =0.04× 60600 × 490 = 1.96.

Therefore

VRV =Vl2 ⇒2RV

RV + 10 = 1.96

⇒ RV =490 Ω.

Voltmeter reading is

VRV =I1RV = 2RVRV + 10 = 2× 490

490 + 10 = 1.96 V.


Current Electricity

5 Figure shows a potentiometercircuit for comparison of tworesistances. The balance point witha standard resistor R = 10.0 Ω isfound to be 58.3 cm, while thatwith the unknown resistance X is68.5 cm. Determine the value ofX . What might you do if youfailed to find a balance point withthe given cell of emf E?Let l1 and l2 denote the balancinglengths obtained with resistances Rand X , separately. HereR = 10.0 Ω, l1 = 58.3 andl2 = 68.5. Let the source of emf Esend a current I into the circuit.

Let k denote the potentialgradient in thepotentiometer wire AB.Then we have forbalancing length l1, thepotential drop across R is

I1R =kl1 (1)


Current Electricity

and for l2, the potential drop across X is

I1X =kl2. (2)

Dividing equation (2) by equation (1), we get

XR = l2

l1⇒ X = l2

l1R

⇒ X =68.558.3 × 10.0 = 1.175× 10

⇒ X =11.75 Ω.

If we fail to find a balance point with the given cell of emf E thenthe potential drop across R and X must be reduced by putting aresistance in series with it. Only if the potential drop across R orX is smaller than the potential drop across the potentiometer wireAB, a balance point is obtained.


Current Electricity

6 Figure shows a 2.0 Vpotentiometer used for thedetermination of internalresistance of a 1.5 V cell. Thebalance point of the cell in opencircuit is 76.3 cm. When aresistor of 9.5Ω is used in theexternal circuit of the cell, thebalance point shifts to 64.8 cmlength of the potentiometerwire. Determine the internalresistance of the cell. GivenE = 1.5 V, l1 = 76.3 cm,R = 9.5 Ω, l2 = 64.8 cm.We know that the internalresistance r of the cell is givenby

r =( l1

l2− 1

)R

⇒ r =(76.3

64.8 − 1)× 9.5

⇒ r =1.68 ΩDr Sukanta Deb Unit II: Current Electricity 47 / 51

Current Electricity

7 Figure shows a potentiometer with a cell of 2.0 V and internalresistance 0.40 Ω maintaining a potential drop across theresistor wire AB. A standard cell which maintains a constantemf of 1.02 V (for very moderate currents upto a few mA)gives a balance point at 67.3 cm length of the wire. To ensurevery low currents drawn from the standard cell, a very highresistance of 600 kΩ is put in series with it, which is shortedclose to the balance point. The standard cell is then replacedby a cell of unknown emf E and the balance point foundsimilarly, turns out to be at 82.3 cm length of the wire.

What is the value E ?What purpose does the high resistance of 600 kΩ have?Is the balance point affected by this high resistance?Is the balance point affected by the internal resistance of thedriver cell?Would the method work in the above situation if the driver cellof the potentiometer had an emf of 1.0 V instead of 2.0 V?Would the circuit work well for determining an extremely smallemf, say of the order of a few mV (such as the typical emf of athermo-couple)? If not, how will you modify the circuit?


Current Electricity

(a) Here E1 = 1.02 V,l1 = 67.3 cm, l2 = 82.3 cm, thenE2 = E =?. We know that

E2E1

= l2l1

⇒ E2 = l2l1E1

⇒ E =82.367.3 × 1.02 = 1.25 V.

(b) The purpose of using thehigh resistance of 600 kΩ is toreduce the current through thegalvanometer when the movablecontact is far from the balancepoint.

(c) The balance point is notaffected by the presence of highresistance.(d) The point is not affected bythe internal resistance of thedriver cell.


Current Electricity

(e) The method would not work if the driver cell of thepotentiometer had an emf of 1.0 V instead of 2.0 V. This isbecause if the emf of the driver cell of the potentiometer is lessthan the emf of the other cell, then there would be no balancepoint on the wire.(f) The circuit would not work well for determining an extremelysmall emf. As the circuit would be unstable, the balance pointwould be close to end A. Hence, there would be a large percentageof error. The given circuit can be modified if a series resistance isconnected with the wire AB. So, the potential drop across AB isslightly greater than the emf measured. So, the percentage errorwould be small.


Current Electricity

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