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Potentiometer and Voltmeter
Construction of a Potentiometer
Potential Gradient
Comparison of emf in a Potentiometer
Summary
A potentiometer is a device that is used to measure the potential difference in a circuit. We know that potential difference is the amount of work done in bringing a charge from one point to other. When there is a potential difference in a circuit, the current flows through the circuit. The unit of potential difference is measured in volts. The potential difference of a circuit can be measured by a voltmeter.
A voltmeter consists of a coil which is suspended between the North Pole and the South Pole of a magnet. So when the current through the circuit the coil deflects and the needle moves on the graduated scale. It also consists of a resistor with high value of resistance which is connected in series with the coil. We are using resistor of high value to minimize the error in the measurement of potential difference. It is because the voltmeter draws a small amount of current from the voltage source of the circuit connected. So the value shown in the voltmeter may not be accurate.
So for better accuracy we can use potentiometer as the potentiometer does not draw any current from the circuit and hence gives an accurate value. A potentiometer is also used to measure the electromotive force of a cell. It is possible to compare the emf of two cell with a potentiometer. It can be used to measure the internal resistance of the cell. The emf is defined as the potential difference between the positive electrode and the negative electrode when there is no current flowing through the cell or in an open circuit. The internal resistance is the resistance that is provided by the electrolyte and electrodes which is present in a cell. Thus it restricts the current flow.
A potentiometer consists of a long wire with uniform area of cross section. Usually the wire is made up of manganin or constantan. In some cases the wire may be cut into some pieces and each piece is connected at the end points by means of a thick metallic strip. Usually it will be copper strips. Each piece of wire has the length of one meter. Usually there will be six pieces of wire and the total length of the wire is six meter. Generally the length of the wire varies from 4 meters to 10 meters. The more the length of the wire, the better the accuracy of the potentiometer.
The potentiometer consists of a driving circuit which consists of a battery, key and rheostat. It also consists of a galvanometer and a jockey. The end points or the terminals of the potentiometer is connected to the points where the potential difference is to be measured.
Potential Gradient is the decrease in potential per unit length. It is calculated as V / L, where V is the potential difference between two points and L is the distance between two points. The longer the wire the lesser the potential gradient and the greater the sensitivity of the potentiometer. Let us see the principle of potentiometer. In the figure we can have a cell with emf E and internal resistance r, a rheostat and the wire. Consider the length of the wire as L and the resistance of the wire as R. This is the primary circuit of the potentiometer. Now consider the point Q along the wire. So the length of the wire from point P to Q is taken as L_{PQ}. The principle of potentiometer states that the potential difference between two points across the potentiometer is directly proportional to the length of the corresponding points.
Thus the voltage across the point P and Q, V_{PQ }∝_{ }L_{PQ}. To remove the proportionality a constant is called which is the potential gradient. It is denoted as K. So V_{PQ} = K L_{PQ}
Potentiometer
For a primary circuit of the potentiometer the potential gradient remains the same.
Potential gradient is calculated as K = V/L, where V is the voltage across the potentiometer wire and the L is the length of the wire in the potentiometer. So the unit of potential gradient is volts/meter. We know that by Ohm’s law Voltage V = I R, where I is current flowing through the circuit and R is the resistance.
So K = V/L = IR/L – equation 1
Now we know that I = V/R
So in this case, Voltage V = E and the total resistance = R + r + Rh, where R is the total resistance of the wire, r is the internal resistance of the cell and Rh is the resistance of the rheostat. So substituting in the equation we get
K = I R/L
= V/R * R/L
= [E/(R + r + Rh)] * R/L
Re arranging the equation we get K = E R/(R + r + Rh) L
The potential gradient can also be written in terms of resistivity. We know that R = ρ L /A, where ρ is the resistivity, L is the length and A is the area of cross section. Substituting the value of R in equation 1 we get
K = (I ρ L/A)/L = I ρ / A.
Problem
Q1. The resistivity of a potentiometer wire is given as 5 * 10^{-6 }Ωm. The area of cross section of the wire is given as 6 * 10^{-4 }m^{2}. Find the potential gradient if a current of 1 A is flowing through the wire.
Solution:
K = V/L
= IR/L
= (IρL/A)/L
= Iρ/A
Substituting the values we get K =1* 5 * 10^{-6}/6 * 10^{-4 }m^{2} = 0.83 * 10^{-2} v/m
Q2. A potentiometer has a wire of length 8 m and the resistance of the wire is 20 Ω. It is connected in series with a cell of emf 2 V and an internal resistance 2 Ω and a rheostat. Find the value of the resistance in rheostat when the potential drop along the wire is 20 µv/mm.
The given values are
L = 8 m, R = 20 Ω, E = 2 V, r = 2 Ω
The potential drop per length mentioned here is the potential gradient.
So K = 20 µv/mm.
We know that K = V/L
= ER/(R + r + Rh)L
So 20 * 10^{-6}/10^{-3} v/m = 2 * 20/(20 + 2 + Rh) 8
= 20 * 10^{-3} = 40/(22 + Rh) 8
20 * 10^{-3 }= 5/22+ Rh
22 + Rh = 5 * 10^{3}/20 = 250
Rh = 250 – 22
= 228 Ω
Consider the potentiometer with the length of the wire L. Two cells E_{1 }and E_{2 }are connected with the help of a two way switch. The switch is then connected to a galvanometer and the other end of the galvanometer is connected to a jockey. The jockey is able to slide on the wire. So now let us connect the cell E_{1 }with the help of the two way switch. By changing the value in the rheostat adjust the reading of the galvanometer to zero. When galvanometer shows zero deflection or null deflection, no current flows through the wire. With the help of jockey let us consider the point of null deflection as A_{1.}
As no current flows through the wire, it is independent of the internal resistance of the cell E_{1. }So the potential difference across the point A A_{1 }= E_{1}. Consider the length of A A_{1 }as L_{1. }So according to the principle of potentiometer we know that E_{1} L_{1. } Thus E_{1} = K L_{1} – equation1, where K is the potential gradient. We know that K = V/L.
Next we will connect the cell E_{2 }with the help of the two way switch. Adjust the reading of the galvanometer to zero by changing the value in the rheostat. We know that no current flows through the wire when the galvanometer shows the null deflection. Let us assume the point of null deflection as A_{2 }with the help of jockey. The potential difference across the point A A_{2} = E_{2.} The length of A A_{1 }is taken as L_{2}. So E_{2} L_{2 }.Thus E_{2} = K L_{2 }.This is the second equation.
Dividing the equation 1 by 2 we get E_{1} = K L_{1}/E_{2} = K L_{2}
Thus E_{1}/E_{2} = L_{1}/L_{2}. Thus we can compare the emf of two cells. We can also use this equation to get the unknown emf.
Q1. A potentiometer of length 1 m has a resistance of 20 Ω. It is then connected with a battery of 8 V and resistor of 5 Ω in series with the wire. Calculate the emf of the primary cell when it gives a balance point at 60 cm.
The length of the wire L = 100 cm
Resistance, R = 20 Ω
Rs = 5 Ω
E = 8 V, L_{1 }= 60 cm
Current I = 8/20 + 5 = 8/25 = 0.32 A
Potential difference across the wire V = IR = 0.32 * 20 = 6.4 V.
Thus according to the principle of potentiometer, E_{2 }/V = L_{1 }/L
E_{2 }= (V/L) * L_{1 }= (6.4/100) * 60
= 3.84 V
To find the internal resistance of the cell
It is also possible to find the internal resistance of the cell with the help of potentiometer.
To find the internal resistance, the cell whose internal resistance is to be measured is placed across a resistance box through a key K_{2}. In the first case the key K_{2 }is opened. Let us assume that J_{ 1 }is the point where the galvanometer shows null deflection. The length of A J_{1} is L_{1. }So E = K * L_{1 –} equation 1. In the second case when the key is closed, the cell gives out a current I through the resistance box R. The point where the galvanometer shows null deflection is taken as J_{2.} The length of A J_{2 }is L_{2.} So the terminal potential difference of the cell is written as V = K L_{2. }This is taken as equation 2. So equation 1 divided by 2 gives
E/V = K L_{1}/K L_{2 - }Equation 3
We know that E = I(R + r) and V = IR
Substituting these values in equation 1 we get, I(R + r)/IR = K L_{1}/K L_{2.}
R + r/R = L_{1}/L_{2}
r = R (L_{1} - L_{2})/L_{2}_{. }Thus we can find the internal resistance of the cell.
A potentiometer is used to measure the potential difference in a circuit. It can also be used to find the emf of a cell and also to compare the emf of two cells. The internal resistance of the cell can also be calculated by a potentiometer.
A voltmeter draws a small amount of current from the voltage source of the circuit connected and thus the reading shown in the voltmeter may not be accurate.
The potentiometer does not draw any current from the circuit and gives an accurate value.
A potentiometer consists of a long wire with uniform area of cross section which is made up of manganin or constantan.
Potential gradient (K) is the potential drop per unit length. It is calculated as V/L, where V is the potential difference between two points and L is the distance between two points. Also K = (IρL/A)/L = Iρ/A.
E_{1}/E_{2} = L_{1}/L_{2} is the equation to compare the emf of two cells, where E_{1} and E_{2} are the emf and L_{1 }and L_{2 }are the length at which it is balanced.
r = R (L_{1} - L_{2})/L_{2} is the formula to find the internal resistance of the cell.
Watch this Video for more reference
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