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What is a Potential Divider? What are they used for?
Description: A wire of uniform resistance is connected to two binding screws A and B [Fig.(a)], A slider or jockey J can move along the wire, remaining in contact with it. J is connected with the binding screw C. This arrangement is generally called potential divider or potentiometer.
Suppose, the length of the resistor AB = L; resistance = R, the length of the portion AJ = 1; its resistance = r.
Since the resistance of the resistor is uniform,
\(\frac{L}{l}\) = \(\frac{R}{r}\) or, r = \(\frac{l}{L}\) ᐧ R ….. (1)
Uses:
Determination of the emf of a cell: Supposer a source of emf E0 is connected between the two ends A and B of the potentiometer wire of resistance R. The cell whose emf E is to be measured is connected between the points A and C through a galvanometer G. It is to be noted that the positive pole of the cell is connected at A [Fig. (b)].
Now, suppose that current in the circuit ACIA due to the effective emf is VAC is i1 and current in that circuit due to the cell = i2. These two currents flow in opposite directions. The slider J is shifted from point to point and placed at a position for which the current through the galvanometer is zero. Clearly at that condition i1 = i2. If R0 be the total resistance of the circuit ACJA, then,
i1 = \(\frac{V_{A C}}{R_0}\) and i2 = \(\frac{E}{R_0}\)
So for zero galvanometer current, \(\frac{V_{A C}}{R_0}\) = \(\frac{E}{R_0}\)
It is evident that current I passes through AJ. So,
VAJ = \(\frac{E_0}{R} \cdot \frac{l}{L} \cdot R\) = \(\frac{l}{L} \cdot E_0\) i.e., VAC = \(\frac{l}{L} E_0\) …. (2)
[length of AB = L, length of AJ = l]
or, E = VAC = \(\frac{l}{L} E_0\) …… (3)
If we already know the values of E0 and L then by noting the value of l from the position of the slider, the emf E of the cell can be determined from the equation (3).
Comparison of the emfs of two cells: From equation (3) we can write, \(\frac{E}{E_0}\) = \(\frac{l}{L}\). So the ratio of the emfs of two cells can be determined with a potentiometer even though the emf of each cell may not be known.
Determination of the internal resistance of a cell: For the length of the potentiometer wire AJ1 = 1 (when the slider is placed at the point J [Fig.]), if galvanometer current is zero, according to the equation (3),
E = \(\frac{l_1}{L} E_0\)
Now, a known resistance R is to be R connected parallel to the cell [Fig.]. Under this condition the potentiometer circuit measures the potential difference V across R. If the galvanometer current is zero for the length of the potentiometer wire AJ2 = l2 (when the slider is placed at J2 [Fig.]),
V = \(\frac{l_2}{L} E_0\) …. (4)
Therefore, \(\frac{V}{E}\) = \(\frac{l_2}{l_1}\) [since, E = \(\frac{l_1}{L} E_0\)]
Again if r be the internal resistance of the cell, E the emf and I be the current in the circuit due to the cell, then
E = I(R + r) and V = IR
i.e., \(\frac{V}{E}\) = \(\frac{R}{R+r}\)
So, \(\frac{l_2}{l_1}\) = \(\frac{R}{R+r}\) or, R + r = \(\frac{l_1}{l_2} \cdot R\) or, r = (\(\frac{l_1}{l_2}\) – 1)R
Therefore, to determine the internal resistance r of the cell, we have no need to know the values of L, E0 or E.
Measurement of the potential difference between two points of an electric circuit: The potential difference between any two points of a circuit can be measured by using potentiometer in the same way as the emf of a cell is measured.
One of the two points (M and N) of an electrical circuit across which the potential difference is to be measured, is connected to the end A of the potentiometer wire and the other point is connected to the jockey J through a galvanometer G [Fig.].
The two poles of the battery of emf E0 are connected to the ends A and B of the potentiometer wire in such a way that the current in the part AM due to the battery of emf E0 and the current i2 due to the external circuit are opposite to each other. Now changing the position of the jockey J from point to point we
observe the position of the jockey for which the galvanometer cur-rent is zero. In this position i1 = i2. Let the total resistance of the circuit MNJA be R0.
Then, i1 = \(\frac{V_{A J}}{R_0}\) [length of AB = L, length of AJ = l]
and i2 = \(\frac{V_{M N}}{R_0}\) So, VMN = VAJ = \(\frac{l}{L} E_0\) …… (5)
If the values of E0 and L are known, then by knowing the value of l from the position of J, the value of VMN i.e., potential difference between M and N can be determined from the equation (5).
Condition of effectiveness: In this arrangement the potentiometer behaves as a voltmeter. So like any voltmeter the resistance of the potentiometer should be very high. So the condition of measurement of potential difference by a potentiometer is that the resistance of the potentiometer wire between the two points A and J must be many times greater than the resistance between the two points M and N of the experimental circuit. If it is not so, then as soon as the potentiometer is connected, the equivalent resistance between the points M and N of the experimental circuit will be reduced to a large extent and much error will be found in the measured value of VMN.
Numerical Examples
Example 1.
The length of the wire of a poentiometer is 100 cm and the emf of a standard cell connected to it is E volt. While measuring the emf of a battery having internal resistance 0.5 Ω the null point is obtained at a length of 30 cm. Determine the emf of the battery. [AIEEE ‘03]
Solution:
When a null point is obtained, no current flows through the battery. So there is no internal potential drop. In this case the internal resistance of the battery has no influence. So if
E’ be the emf of the battery, then
\(\frac{E}{100}\) = \(\frac{E^{\prime}}{30}\) or, \(\frac{E}{E^{\prime}}\) = \(\frac{10}{3}\) or, E’ = 0.3E.
Example 2.
A potential difference of 220 V is applied at the two ends of a rheostat of 12000 Ω [Fig.]. A voltmeter of resistance 6000 Ω is connected between the points A and D. If the point D divides AB in the ratio of 1 : 4 what will be the reading of the voltmeter?
Solution:
Since the voltmeter is connected between the points A and D, its reading will be 33.3V.
Example 3.
In a potentiometric arrangement, a cell is connected to the potentiometer wire of 60 cm in length to make the deflection zero in galvanometer. Now if the cell is shunted with a 6 Ω resistor, a null point is found in 50 cm length of the wire. What is the internal resistance of the cell? [AIEEE ‘03]
Solution:
Here, shunt R = 6 Ω, l1 = 60 cm, l2 = 50 cm.
∴ Internal resistance of the cell,
r = \(\left(\frac{l_1}{l_2}-1\right) R\) = \(\left(\frac{60}{50}-1\right)\) × 6 = 0.2 × 6 = 1.2Ω
Example 4.
A length of potentiometer wire of 155 cm balances the emf of a cell in a circuit and a length of 135 cm when the cell has a conductor of resistance 8 Ω connected between its terminals. Find the internal resistance of the cell. [WBCHSE Sample Question]
Solution:
Let the emf of the cell E, and its internal resistance = r. When a resistance R is connected between its terminals (here, R = 8Ω), the cell is subjected to an internal drop of potential. Then, the terminal potential difference,