Physics WHEATSTONE BRIDGE ,METER BRIDGE, FOR CBSE-NCERT-6
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WHEATSTONE BRIDGE ,METER BRIDGE, FOR CBSE-NCERT-6

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`color{blue}{star}` WHEATSTONE BRIDGE
`color{blue}{star}` METER BRIDGE
`color{blue}{star}` POTENTIOMETER

WHEATSTONE BRIDGE

`color{blue} ✍️` As an application of Kirchhoff’s rules consider the circuit shown in Fig. 3.25, which is called the Wheatstone bridge.

`color{blue} ✍️` The bridge has four resistors `color{blue}(R_1, R_2, R_3)` and `R_4`. Across one pair of diagonally opposite points (A and C in the figure) a source is connected. This (i.e., AC) is called the battery arm.

`color{blue} ✍️` Between the other two vertices, B and D, a galvanometer G (which is a device to detect currents) is connected. This line, shown as `BD` in the figure, is called the galvanometer arm.

`color{blue} ✍️` For simplicity, we assume that the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current `I_g` through `G`. Of special interest, is the case of a balanced bridge where the resistors are such that `I_g = 0`.

`color{blue} ✍️` We can easily get the balance condition, such that there is no current through `G`. In this case, the Kirchhoff’s junction rule applied to junctions `D` and ` B` (see the figure) immediately gives us the relations `color{blue}(I_1 = I_3)` and `color{blue}(I_2 = I_4)`.

`color{blue} ✍️` Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives

`color{blue}(–I_1 R_1 + 0 + I_2 R_2 = 0 \ \ \ \ (Ig = 0))`

.........(3.81)

`=>` and the second loop gives, upon using `color{blue}(I_3 = I_1, I_4 = I_2)`

`color{blue}(I_2 R_4 + 0 – I_1 R_3 = 0)`

...........(3.82)

From Eq. (3.81), we obtain, `color{blue}((I_1)/(I_2) = (R_2)/(R_1))`

whereas from Eq. (3.82), we obtain, `color{blue}((I_1)/(I_2) = (R_4)/(R_3))`

Hence, we obtain the condition

`color{blue}((R_2)/(R_1) = (R_4)/(R_3))`

...........[3.83(a)]

`color{blue} ✍️` This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection.

`color{blue} ✍️` The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance.

`color{blue} ✍️` Let us suppose we have an unknown resistance, which we insert in the fourth arm; `R_4` is thus not known. Keeping known resistances `R_1` and `R_2` in the first and second arm of the bridge, we go on varying `R_3` till the galvanometer shows a null deflection.

`color{blue} ✍️` The bridge then is balanced, and from the balance condition the value of the unknown resistance `R_4` is given by

`color{blue}(R_4 = R_3 (R_2)/(R_1))`

.........[3.83(b)]

`color{blue} ✍️` A practical device using this principle is called the meter bridge. It will be discussed in the next section.
Q 3174191956

The four arms of a Wheatstone bridge (Fig. 3.26) have the following resistances:
`AB = 100Ω, BC = 10Ω, CD = 5Ω, and DA = 60Ω.`
A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of `10 V` is maintained across AC.
Class 12 Chapter example 8
Solution:

Considering the mesh BADB, we have
`100I_1 + 15I_g – 60I_2 = 0`

or `20I_1 + 3I_g – 12I_2= 0`

Considering the mesh BCDB, we have

`10 (I_1 – I_g) – 15I_g – 5 (I_2 + I_g) = 0`
`10I_1 – 30I_g –5I_2 = 0`
`2I_1 – 6I_g – I_2 = 0`

Considering the mesh ADCEA,

`60I_2 + 5 (I_2 + I_g) = 10`

`65I_2 + 5I_g = 10`

`13I_2 + I_g = 2`
Multiplying Eq. (3.84b) by 10

`20I_1 – 60I_g – 10I_2 = 0`

From Eqs. (3.84d) and (3.84a) we have

`63I_g – 2I_2 = 0`
`I_2 = 31.5I_g`

Substituting the value of `I_2` into Eq. [3.84(c)], we get
`13 (31.5I_g ) + I_g = 2`
`410.5 I_g = 2`
`I_g = 4.87 mA.`

METER BRIDGE

`color{blue} ✍️` The meter bridge is shown in Fig. 3.27. It consists of a wire of length `1m` and of uniform cross sectional area stretched taut and clamped between two thick metallic strips bent at right angles, as shown.

`color{blue} ✍️` The metallic strip has two gaps across which resistors can be connected. The end points where the wire is clamped are connected to a cell through a key. One end of a galvanometer is connected to the metallic strip midway between the two gaps.

`color{blue} ✍️` The other end of the galvanometer is connected to a ‘jockey’. The jockey is essentially a metallic rod whose one end has a knife-edge which can slide over the wire to make electrical connection.

`color{blue} ✍️` R is an unknown resistance whose value we want to determine. It is connected across one of the gaps. Across the other gap, we connect a standard known resistance S. The jockey is connected to some point D on the wire, a distance l cm from the end A.

`color{blue} ✍️` The jockey can be moved along the wire. The portion AD of the wire has a resistance `R_(cm)l,` where `R_(cm)` is the resistance of the wire per unit centimetre.

`color{blue} ✍️` The portion DC of the wire similarly has a resistance `color{blue}(R_(cm) (100-l))`. The four arms AB, BC, DA and CD [with resistances `R, S, R_(cm) l` and `color{blue}(R_(cm)(100-l))` obviously form a Wheatstone bridge with AC as the battery arm and BD the galvanometer arm.

`color{blue} ✍️` If the jockey is moved along the wire, then there will be one position where the galvanometer will show no current.

`color{blue} ✍️` Let the distance of the jockey from the end A at the balance point be `l= l_1`. The four resistances of the bridge at the balance point then are `color{blue}(R, S, R_(cm) l_1)` and `color{blue}(R_(cm)(100–l_1).)` The balance condition, Eq. [3.83(a)] gives

`color{blue}(R/S = (R_(cm)l_1)/(R_(cm)(100-l_1)) = (l_1)/(100-l_1))`

.........(3.85)

`color{blue} ✍️` Thus, once we have found out `l_1`, the unknown resistance R is known in terms of the standard known resistance S by

`color{blue}(R = S (l_1)/(100 - l_1))`

..........(3.86)

`color{blue} ✍️` By choosing various values of S, we would get various values of `l_1,` and calculate R each time. An error in measurement of `l_1` would naturally result in an error in R. It can be shown that the percentage error in R can be minimised by adjusting the balance point near the middle of the bridge, i.e., when `l_1` is close to 50 cm. (This requires a suitable choice of S.)
Q 3115101060

In a metre bridge (Fig. 3.27), the null point is found at a distance of 33.7 cm from A. If now a resistance of `12Ω` is connected in parallel with
Class 12 Chapter example 9
Solution:

From the first balance point, we get

`R/S = (33.7)/(66.3)`

After S is connected in parallel with a resistance of `12Ω` , the resistance across the gap changes from `S` to `S_(eq,)` where

`S_(sq) = (12S)/(S+12)`

and hence the new balance condition now gives

`(51.9)/(48.1) = R/S_(Sq) = R((S+12)/(12S))`

Substituting the value of R/S from Eq. (3.87), we get

`(51.9)/(48.1) = (S+12)/(12) • (33.7)/(66.3)`

which gives `S = 13.5Ω.` Using the value of R/S above, we get `R = 6.86 Ω.`

POTENTIOMETER

`color{blue} ✍️` This is a versatile instrument. It is basically a long piece of uniform wire, sometimes a few meters in length across which a standard cell is connected.

`color{blue} ✍️` In actual design, the wire is sometimes cut in several pieces placed side by side and connected at the ends by thick metal strip. (Fig. 3.28). In the figure, the wires run from A to C.



`color{blue} ✍️` The small vertical portions are the thick metal strips connecting the various sections of the wire. A current I flows through the wire which can be varied by a variable resistance (rheostat, R) in the circuit. Since the wire is uniform, the potential difference between A and any point at a distance l from A is

`epsilon(l) = phil`

..........(3.89)

`color {blue}{➢➢}` where `φ` is the potential drop per unit length. Figure 3.28 (a) shows an application of the potentiometer to compare the emf of two cells of emf `ε_1` and `ε_2 `.

`color{blue} ✍️` The points marked 1, 2, 3 form a two way key. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to `ε_1.`

`color{blue} ✍️` The jockey is moved along the wire till at a point `N_1,` at a distance `l_1` from A, there is no deflection in the galvanometer. We can apply Kirchhoff’s loop rule to the closed loop `color{blue}(AN_1G31A)` and get,

`color{blue}(phi l_1 + 0 - epsilon_1 =0)`

...........(3.90)

`color {blue}{➢➢}`Similarly, if another emf `ε_2` is balanced against `color{blue}(l_2 (AN_2))`

`color{blue}(phi l_2 + 0 - epsilon_2=0)`

.........(3.91)

From the last two equations

`color{blue}((epsilon_1)/(epsilon_2) = (l_1)/(l_2))`

..........(3.92)

`color{blue} ✍️` This simple mechanism thus allows one to compare the emf’s of any two sources.

`color{blue} ✍️` In practice one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then easily calculated from Eq. (3.92). We can also use a potentiometer to measure internal resistance of a cell [Fig. 3.28 (b)].

`color{blue} ✍️` For this the cell (emf ε ) whose internal resistance (r) is to be determined is connected across a resistance box through a key `K_2,` as shown in the figure. With key `K_2` open, balance is obtained at length `l_1 (AN_1).` Then,

`color{blue}(epsilon= phil_1)`

.....[3.93(a)]

`color{blue} ✍️` When key `K_2` is closed, the cell sends a current `(I )` through the resistance box `(R)`. If `V` is the terminal potential difference of the cell and balance is obtained at length `color{blue}(L_2 (AN_2)),`


`color{blue}(V = phil_2)`

...........[3.93(b)]

`color{blue} ✍️` So we have

`color{blue}(ε//V= l_1//l_2)`

............[3.94(a)]

But, `color{blue}(ε = I (r + R))` and `color{blue}(V = IR).` This gives

`color{blue}(ε//V = (r+R)//R)`

...........[3.94(b)]
From Eq. [3.94(a)] and [3.94(b)] we have

`color{blue}((R+r)//R = l_1//l_2)`

`color{blue}(r = R (l_1/l_2-1)`

..............(3.95)

`color{blue} ✍️` Using Eq. (3.95) we can find the internal resistance of a given cell. The potentiometer has the advantage that it draws no current from the voltage source being measured. As such it is unaffected by the internal resistance of the source
Q 3155101064

A resistance of `R Ω` draws current from a potentiometer. The potentiometer has a total resistance `R_0 Ω` (Fig. 3.29). A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.
Class 12 Chapter example 10
Solution:

While the slide is in the middle of the potentiometer only half of its resistance `(R_0//2)` will be between the points A and B. Hence, the total resistance between A and B, say, `R_1`, will be given by the following expression:



`(1)/(R_1) = 1/R + (1)/(R_0/2)`

`R_1 = (R_0R)/(R_0+2R)`


The total resistance between A and C will be sum of resistance between A and B and B and C, i.e., `R_1 + R_0//2`

∴ The current flowing through the potentiometer will be

`I = (V)/(R_1+R_0//2) = (2V)/(2R_1+R_0)`

The voltage `V_1` taken from the potentiometer will be the product of
current `I` and resistance `R_1,`

`V_1 = IR_I = ((2V)/(2R_1+R_0))xxR_1`

Substituting for `R_1,` we have a

`V_1 =((2V)/((R_0xxR)/(R_0+2R))) xx (R_0xxR)/(R_0+2R)`

`V_1 = (2VR)/(2R+R_0+2R)`

or `V_1 = (2VR)/(R_0+4R)`
 
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