why the reading of EMF by a voltmeter is less than that by a potentiometer
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Answered by
4
We know that the terminal voltage of a cell is the potential difference between its electrodes.
A voltmeter should not be used to measure the emf of a cell as a voltmeter draws some current from the cell during measurement. When voltmeter is connected across the cell it amounts to closed circuit measurements and there is inner potential drop due to internal resistance of the cell. Therefore, the measurement of cell's emf will not be accurate
To measure a cell's emf, use of a potentiometer is preferred since in potentiometer-measurement no current flows through the cell. It is an open circuit measurements. The emf of cell will be measured accurately.
Other advantages of using a potentiometer are
Sensitivity of potentiometer is high as compared to that of voltmeter.
The resistance of potentiometer becomes infinite at the time of measurement. Whereas in case of voltmeter the resistance is very high but still a measurable number.
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Look at the figure below

http://dmr-physicsnotes.blogspot.in
A potentiometer is a circuit that is used to measure EMF of a cell. the circuit is as shown: A battery of known emf
ε
is connected with the end terminals A and B of potentiometer with rheostat
Rh
, ammeter, resistance box and key K all in series.
The ends of a resistance
R1
are connected to terminal A and jockey J through galvanometer. The resistance
R1
is connected to cell
E1
and key
K1
as shown.
Step 1. Close key K. Adjust the resistance
R
from resistance box so that the potential difference across the potentiometer wire is greater than the EMF of the cell, which is to be measured.
Suppose
RP
is the resistance of the potentiometer wire of length
=L
(100cm for a lab model Meter potentiometer and Length of the potentiometer wire varies from 5 to 10m).
Current through the potentiometer
I=εR+Rp
Potential drop across potentiometer wire
=εR+Rp×Rp
.....(1)
Step 2. Close key
K1
. Current flows through resistance
R1
due to which potential difference is developed across this resistance.
Step 3. Adjust the position of jockey on potentiometer wire such that there is no deflection seen on the galvanometer. Indicates a balanced circuit and that the potentiometer is another null- measurement device.
Let this position of jockey be is J.
Let length AJ of potentiometer wire be
=l
.
Let
K
be the potential gradient of potentiometer wire. From (1)
K=ε
R+
Rp×
Rp
L
.........(2)
With Galvanometer showing no deflection, EMF of the cell is equal to the potential difference
V
across
R1
. Now
V=K×l
......(3).
Inserting value of
K
from (2)
V=ε×RpR
A voltmeter should not be used to measure the emf of a cell as a voltmeter draws some current from the cell during measurement. When voltmeter is connected across the cell it amounts to closed circuit measurements and there is inner potential drop due to internal resistance of the cell. Therefore, the measurement of cell's emf will not be accurate
To measure a cell's emf, use of a potentiometer is preferred since in potentiometer-measurement no current flows through the cell. It is an open circuit measurements. The emf of cell will be measured accurately.
Other advantages of using a potentiometer are
Sensitivity of potentiometer is high as compared to that of voltmeter.
The resistance of potentiometer becomes infinite at the time of measurement. Whereas in case of voltmeter the resistance is very high but still a measurable number.
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.-.-.-
Look at the figure below

http://dmr-physicsnotes.blogspot.in
A potentiometer is a circuit that is used to measure EMF of a cell. the circuit is as shown: A battery of known emf
ε
is connected with the end terminals A and B of potentiometer with rheostat
Rh
, ammeter, resistance box and key K all in series.
The ends of a resistance
R1
are connected to terminal A and jockey J through galvanometer. The resistance
R1
is connected to cell
E1
and key
K1
as shown.
Step 1. Close key K. Adjust the resistance
R
from resistance box so that the potential difference across the potentiometer wire is greater than the EMF of the cell, which is to be measured.
Suppose
RP
is the resistance of the potentiometer wire of length
=L
(100cm for a lab model Meter potentiometer and Length of the potentiometer wire varies from 5 to 10m).
Current through the potentiometer
I=εR+Rp
Potential drop across potentiometer wire
=εR+Rp×Rp
.....(1)
Step 2. Close key
K1
. Current flows through resistance
R1
due to which potential difference is developed across this resistance.
Step 3. Adjust the position of jockey on potentiometer wire such that there is no deflection seen on the galvanometer. Indicates a balanced circuit and that the potentiometer is another null- measurement device.
Let this position of jockey be is J.
Let length AJ of potentiometer wire be
=l
.
Let
K
be the potential gradient of potentiometer wire. From (1)
K=ε
R+
Rp×
Rp
L
.........(2)
With Galvanometer showing no deflection, EMF of the cell is equal to the potential difference
V
across
R1
. Now
V=K×l
......(3).
Inserting value of
K
from (2)
V=ε×RpR
Answered by
20
Because when we use potentiometer the current flow is zero then no voltage is dropped.but,in the case of voltmeter the current is flow and some voltage will be dropped.so,reading of emf by voltmeter is less than that of potentiometer
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