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Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

(i) Wheatstone designed a network of four resistances with the help of which the resistance of a given conductor can be measured. Such a network of resistances is known as Wheastone's bridge.

(ii) In this bridge, four resistance P, Q, R and S are so connected so as to form a quadrilateral ABCD. A sensitive galvanometer and key K2 are connected between diagonally opposite corners B and D, and a cell and key K1 are connected between two other corners A and C (figure shown)

(iii) When key K1 is pressed, a current i flows from the cell. On reaching the junction A, the current i gets divided into two parts i1 and i2. Current i1 flows in the arm AB while i2 in arm AD. Current i1, on reaching the junction B gets further divided into two parts (i- ig) and ig, along branches BC and BD respectively. At junction D, currents i2 and ig are added to give a current (i2  ig), along branch DC. (i2 - ig) and (i2  ig) add up at junction C to give a current (i1  i2) or i along branch CE. In this way, currents are distributed in the different branches of bridge. In this position, we get a deflection in the galvanometer.

(iv) Now the resistance P,Q,R and S are so adjusted that on pressing the key K2, deflection in the galvanometer becomes zero or current ig in the branch BD becomes zero. In this situation, the bridge is said to be balanced.

(v) In this balanced position of bridge, same current i1 flows in arms AB and BC and similarly same current i2 in arms AD and DC. In other words, resistances P and Q and similarly R and S, will now be joined in series.

(vi) Condition of balance : Applying Kirchhoff's 2nd law to mesh ABDA, i1P + igG - i2R = 0 ...(1)

Similarly, for the closed mesh BCDB, we get, (i1 - ig) Q - (i2 + ig)S - igG = 0 ...(2) When bridge is balanced, ig = 0. Hence eq. (1) & (2) reduce to
i1P - i2R = 0 or i1P = i2R ....(3)
i1Q - i2S = 0 or i1Q = i2S ...(4)
Dividing (3) by (4), we have, Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET ....(5)
This is called as condition of balanced for Wheatstone's Bridge.

(vii) It is clear from above equation that if ratio of the resistance P and Q, and the resistance R are known, then unknown resistance S can be determined. This is the reason that arms P and Q are called as ratio arms, arm AD as known arm and arm CD as unknown arm.

(viii) When the bridge is balanced then on inter-changing the positions of the galvanometer and the cell there is no effect on the balance of the bridge. Hence the arms BD and AC are called as conjugate arms of the bridge.

(ix) The sensitivity of the bridge depends upon the value of the resistances. The sensitivity of bridge is maximum when all the four resistances are of the same order.

Ex.22 Find equivalent resistance of the circuit between the terminals A and B.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Sol. Since the given circuit is wheat stone bridge and it is in balance condition.
Because, 10 × 3 =30 = 6× 5

hence this is equivalent to

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Ex.23 Find the equivalent resistance between A and B

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET
Sol. This arrangement can be modified as shown in figure since it is balanced wheat stone bridge.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

8.1 Unbalanced Wheatstone Bridge

Ex.24 

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Find equivalent resistance ?
Sol. Let potential at point B is x and E is Y

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Applying KCl at point B

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

8x - 5y = v ...(1)

Applying KCL at point E

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

⇒ 8y - 5x = 2v ...(2)

solving x & y Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET, Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

current from branches BC & EF adds up to give total current (i) flowing in the circuit.

i = i3  i4 = Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET = Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Because,  Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET          Therefore, Req. = Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Ladder Problem :

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Find the effective resistance between A & B ?

Sol. Let the effective resistance between A & B be RE since the network is infinite long, removal of one cell from the chain will not change the network. The effective resistance between points C & D would also be RE.
The equivalent network will be as shown below

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

The original infinite chain is equivalent to R in series with R & RE in parallel.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

RER + RE2 = R2 + 2RRE ⇒ RE2 - RRE - R2 = 0

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Ex.25 

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Find the equivalent resistance between A & B ?

Sol. As moving from one section to next one, resistance is increasing by k times. Since the network is infinitely long, removal of one section from the chain will bring a little change in the network. The effective resistance between points C & D would be kRE (where RE is the effective resistance)

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Therefore, Effective R between A & B.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

On solving we get

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

15. Symmetrical Circuits :

Some circuits can be modified to have simpler solution by using symmetry if they are solved by traditional method of KVL and KCL then it would take much time.

Ex.26 Find the equivalent Resistance between A and B

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Sol. I Method : mirror symmetry

The branches AC and AD are symmetrical

Therefore, current through them will be same.

The circuit is also similar from left side and right side like mirror images with a mirror placed alone CD therefore current distribution while entering through B and an exiting from A will be same. Using all these facts the currents are as shown in the figure. It is clear that current in resistor between C and E is 0 and also in ED is 0. It's equivalent is shown in figure (b)

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

II Method : folding symmetry

Therefore,  The potential difference in R between (B, C) and between (B, D) is same VC = VD

Hence the point C and D are same hence circuit can be simplified as

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

This is called folding.

Now , it is Balanced Wheatstone bridge

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET In II Method it is not necessary to know the currents in CA and DA

Ex.27 Find the equivalent Resistance between A and B

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Sol. In this case the circuit has symmetry in the two branches AC and AD at the input

Therefore, current in them are same but from input and from exit the circuit is not similar

(Because, on left R and on right 2R)

Therefore, on both sides the distribution of current will not be similar.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Here Vc = Vd

hence C and D are same point
So, the circuit can be simplified as

Now it is balanced wheat stone bridge.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET = Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Ex.28 Find the equivalent Resistance between A and B

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Sol. Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Here VA = VC and VB = VD

Here the circuit can be simplified as

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET = Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET 

Ans. Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Ex.29 Twelve equal resistors each R W are connected to form the edges of a cube. Find the equivalent resistances of the network.

(a) When current enters at 1 & leaves at 6 (body diagonal)

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET
Sol. Here 2, 4, 8 are equipotential points (if we move from 1 → 2, 4, 8 it comes along the edge & 6 → 2, 4, 8 it comes along face diagonal). Similarly 3, 5, 7 are equipotential points.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET
(b) When current enters at 1 and leaves at 2

Sol. Here 3, 7 are equipotential surface (if we move from 1 → 3, 7 we have along face and 2, →3, 7 we move along edge) similarly 4, 8 are equipotential surface.

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET
(c) When current enters at 1 and leaves at 3

Sol. If we cut the cube along the plane passing through 2, 4, 5, 7 then by mirror symmetry, the final configuration will be

Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

Meter Bridge
It is based on principle of whetstone bridge. It is used to find out unknown resistance of wire. AC is 1 m long uniform wire R.B. is known resistance and S is unknown resistance. A cell is connected across 1 m long wire and Galvanometer is connected between Jockey and midpoint D. To find out unknown resistance we touch jockey from A to C and find balance condition. Let balance is at B point on wire.
Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

AB = ℓ cm  P = r ℓ

BC = (100 – ℓ) cm Q = r(100 – ℓ) where r = resistance per unit length on wire.

At balance condition :
Wheatstone Bridge & Meter Bridge | Physics Class 12 - NEET

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FAQs on Wheatstone Bridge & Meter Bridge - Physics Class 12 - NEET

1. What is a Wheatstone bridge and how does it work?
Ans. A Wheatstone bridge is a circuit used to measure unknown electrical resistances. It consists of four resistors arranged in a diamond shape, with a voltage source connected to two opposite corners and a galvanometer connected to the other two corners. When the resistance of the unknown component is adjusted until the galvanometer reads zero, the bridge is said to be balanced.
2. How can a Wheatstone bridge be used to measure an unknown resistance?
Ans. To measure an unknown resistance using a Wheatstone bridge, the bridge is first balanced by adjusting the known resistances. Once the bridge is balanced, the ratio of the two known resistances is equal to the ratio of the unknown resistance to the fourth resistance. By knowing the values of the known resistances and measuring the fourth resistance, the value of the unknown resistance can be determined.
3. What is a meter bridge and what is its purpose?
Ans. A meter bridge, also known as a slide wire bridge, is a device used to measure unknown resistances. It consists of a long wire with a uniform cross-section, a sliding contact called a jockey, and a galvanometer. The unknown resistance is connected in series with a known resistance, and the jockey is moved along the wire until the galvanometer reads zero. The length of the wire between the jockey and the known resistance can then be used to calculate the value of the unknown resistance.
4. How does a meter bridge work?
Ans. A meter bridge works based on the principle of Wheatstone bridge. When the jockey is moved along the wire, the resistance between the jockey and the known resistance changes. By moving the jockey until the galvanometer reads zero, the ratio of the resistance between the jockey and the known resistance is equal to the ratio of the unknown resistance to the known resistance. This allows for the calculation of the unknown resistance value.
5. How accurate are the measurements obtained using a meter bridge?
Ans. The accuracy of the measurements obtained using a meter bridge depends on various factors such as the quality of the wire, the precision of the jockey movement, and the sensitivity of the galvanometer. However, meter bridges are generally considered to provide reasonably accurate measurements, especially when used with high-quality components and careful calibration. The accuracy can also be improved by repeating the measurements multiple times and taking an average value.
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