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To determine the internal resistance of a cell by a potentiometer. Theory: The The pote potent ntio iome mete terr prim primar aril ily y meas measur ures es potential difference. The potentiometer circui circuitt is shown shown in the figur figuree – 1. AB is a length of uniform un iform resistive wire. Terminals Terminals X and Y are connect connected ed across across the potent potential ial difference (V1) being measured in the same way as those of a voltmeter vo ltmeter would be.
If the positive terminal of the driving cell is connect connected ed to X (as shown in the diagram), then X must be connected to the positive side of the the pote potent ntiial dif differen erence ce (V1) being measured. The The pote potent ntiom iomet eter er is said said to be balanced when the jocey (sliding contact) is at such a position on !" that there is no curr current ent through the galvanometer. !t balance, I# $ %, and therefore I = I1&(by applying irchhoff's 1st law at C)
!nd so I = ! &(by applying irchhoff's 1st law at A)
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Therefore,
*otential at X
$ *otential at A
(since I $ %)
!nd,
*otential at Y
$ *otential at C
(since I# $ %)
Therefore, *otential difference between X and Y $ *otential difference between A and C i.e. V 1 = I (R1 ) " where # 1 is resistance of wire between points A and C. σ
=
R1 L1
+ow, let
& where $1 is length of wire between points A and C. V 1
= I σ L
1
Therefore we have .(1) If V 1 is replaced by another potential difference V 2 and a new balance length L2 is found, then V (
= I σ L
(
.() V 1 V (
-ividing euation (1) by euation () gives
=
L1 L(
..()
Thus, by measuring the respective balance lengths, two potential differences can be compared. !t balances, I# $I $ %, and therefore no current is drawn from the potential differences being compared. /ell or any other source which supplies a potential difference to the circuit to which it is connected, has within it some resistance called internal resistance. 0hen there is no current in the cell i.e., at open circuit, the potential difference between its terminal is ma2imum and is called its electromotive force (e.m.f.). 0hen the cell is discharging i.e., at closed circuit, its terminal potential difference between its terminal is reduced to V because of internal drop of potential across its internal resistance r . #
K 2 E 1
r
#
A K 1
DC Power Supply
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B
J
3 %i&ure' 4 5inding internal resistance of a cell by a potentiometer.
In the above figure the balance point for the cell E 1 whose internal resistance r is to be determined , is found out as usual , at a distance L1 from the end A of the potentiometer with the ey ( open .
= I σ L
E 1
1
.(3) Then the resistance # is introduced in the resistance bo2 and the ey ( is closed , now i current is flowing through the circuit ( through e2ternal resistances R and internal resistance r ). the potential difference between the terminals of the cell )1 falls because of internal drop of potential across its internal resistance r .
∴
E 1 $ *otential difference between the terminals 6 lost voltage in the cell
∴ E 1 =
iR + ir = V + ir
!s E 1 and V are the potential differences between the terminals of the cell at the open and closed circuits. ! balance point is now found at a distance L2 from the end A of the potentiometer and V is the potential difference between the terminals of the cell at closed circuit. Therefore V
= iR = I σ L
(
.(7) from euation (3) and (7) we have E 1 V
=
L1
iR
L(
+ ir
iR
=
L1 L(
or
r
= L − 1 R L 1
(
i.e. .(8) "y nowing L1 , L2 and R we can determine internal resistance of a cell by euation (8)
Apparatus : *otentiometer, wires.
-/ power supply, /ell 1 , 9esistance bo2 , 9heostat (9h) , #alvanometer , /onnecting
Connection : /onnect the positive terminal of the *C po+er supply to the scre+ A ,end A) of the potentiometer and the negative terminal of the -/ power supply through a rheostat (9h) to the scre+ B ,end B- of the potentiometer with a ey K 1 (crocodile clip of the wire can be as ey K 1, when clip is connected K 1 (circuit) is closed). :oin the positive terminal of the cell E 1 to the scre+ A of the potentiometer and negative terminal of the cell 1 through the &alanometer / to the 0ocey 2 . !lso connect the resistance bo3 ( R) through a ey K 2 to the two terminals of the cell )1. 4rocedure: 1. !djust a small resistance in the rheostat #h and close the ey ( 1 by connecting crocodile clip to the -/ power supply. ;eep the ey ( open,do not connect resistance # +ith battery at the be&innin&-
7 .
!t first press the jocey near point the end A and then near the end B of the potentiometer wire. If the galvanometer deflections are in the same direction, then either the resistance in the #h is too &reat or olta&e of the *C po+er supply is too small . *ecrease the resistance in #h or increase the voltage in the *C po+er supply until the opposite deflections are obtained at the two ends A and B. !djustment of #h should be such that as to get a null point on the 5th or 1!th +ire.
. +ow find out the balance point accurately when ( 1 is closed and ( is open .
*iscussions: 1) The internal resistance of a cell depends on the strength of the current. It decreases as the current increases. It is , therefore, better to change the e2ternal resistance over a range of 3% ohms and then to calculate the mean values of r. ) !fter every reading, the ey ; 1 should be opened to allow the wire to cool. ) /are should be taen to see that ; is open when determining l 2. 3) The internal resistance of a cell can be determined with voltmeter and ammeter also but this method is more accurate.
*ata sheet: #esistance
>ean of
#esistance
In
=alue of
In
=alue
#
L1
L1
#
L2
of the cell
ohms
cm
cm
ohms
cm
ohms
Internal resistance
>ean
r
r
L r = L − 1 R 1
(
1! Both (ey
;ey
! ( 1 Closed
and (
( 1 and infinity
7!
( closed
8!
ohms
8
open
9!
Calculation:
r
L = − 1 R L 1
(
$ .. )rror calculation: n
∑ ( r − r )
(
i
∆r =
=
i 1
n
−1 $
#esults : r Internal resistance of the cell,
± ∆r = $ .ohms ample oral ;uestions and Ans+ers )3periment
1.
0hat do you understand by the internal resistance of a cell? Ans: 0hen the e2ternal circuit is complete, within the cell (or battery) a current flow from the plate at a lower potential to the plate at higher potential and the medium between the plates offers a resistance to the flow of current. This resistance is nown as the internal resistance of the cell.
.
A current where as the e.m.f. of the circuit is constant both in magnitude and direction. .>.5. is present both in the open and closed circuit while potential difference is present in the closed circuit. 9.
If even with correct connections, deflections on the first and last wires are in same direction for the cell, what would you do? Ans : The current sent by the driver cell is such that the total potential drop across the wire is greater than the e.m.f. of one cell but less than the e.m.f. of the other cell. The potentiometer current is to be increased by diminishing the rheostat resistance.
. -oes the resistance of the galvanometer interfere with the p osition of the null point? Ans : +o, it saves the galvanometer from damage. >. @ow can you increase the sensitiveness of the potentiometer? Ans : "y increasing the balancing length of the potentiometer as long as possible. ?. 0hat ind of wire would you use as the potentiometer wire and why? Ans : The material should be so chosen that it may have high resistance ( such as e#re$a , mananin , etc.).
If the wire be heated by the passage of current, would the null point change? Ans : Bes , since the resistance of the wire changes with the change of temperature. @ence the potentiometer circuit should be closed only for the time during which the null point is determined.
*otential at X (since I $ %)
$
*otential at Y (since I# $ %)
*otential at A
$
"
*otential at C "
Therefore, *otential difference between X and Y $ *otential difference between A and C i.e. V 1 = I (R1 ) " 0here # 1 is resistance of wire between points A and C. σ
=
R1 L1
+ow, let & 0here $1 is length of wire between points A and C. Therefore we have V 1
= I σ L
1
.(1)
C
D
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1. If ne&atie terminal of the -/ power supply (-riving cell) in the diagram below is connected to point A, then will the null point occur? 0hy? Ans4 +o. . !fter getting one null point, the circuit should be left open for E minutes before the determination of the ne2t null point is taen up. 0hy? Ans4 9esistance changes. 7. If resistance of the rheostat ,# h- is increased and others are ept constant, then where does the null point occur& closer or further from the point A@ hy@ V 1
= I σ L
1
Ans: 5urther from the point A. e no+ decreased. !s =1 is constant therefore F1 is increased.
. 0hen # h is increased current I is
If -/ po+er supply olta&e (-riving battery) is smaller than that of cell voltage ( )1- . 0ill the null point occur? 0hy? !ns4 +o 9. If voltage of the *C po+er supply olta&e (-riving cell) is increased and others are ept constant, then where does the null point occur& closer or further from the point A@ hy@ 8.
V 1
= I σ L
1
. 0hen -/ po+er supply olta&e (-riving Ans: /loser to the point A. e no+ cell) is increased current I is increased. !s =1 is constant therefore F1 is decreased. . If resistance # is connected to the cell )1 then where does the null point occur& closer or further from the point A@ hy@ V 1
= I σ L
Ans: Closer to the point A. e no+ constant therefore F1 is decreased. 4re'$ab )3ercise:
1. 0hat would you e2pect the voltmeter in the adjacent circuit to do when the potentiometer wiper is moved to the right? ........................................................................................ ........................................................................................
1
. 0hen =1 is decreased and current I is
1% . 0hich way would you have to move the potentiometer wiper, to the left or to the right, in order to increase current through resistor 91 in the adjacent circuit?
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. 0hat is the difference between e.m.f and potential difference? ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................
3. 0hy does internal resistance of a battery occur? ............................................................................................................................................................................ ............................................................................................................................................................................
9. If resistance of the rheostat ,# h- is increased and others are ept constant in the adjacent circuit, then where does the null point occur& closer or further from the point A@ !nd why@
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