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(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 K_{2} are connected between diagonally opposite corners B and D, and a cell and key K_{1} are connected between two other corners A and C (figure shown)

(iii) When key K_{1} is pressed, a current i flows from the cell. On reaching the junction A, the current i gets divided into two parts i_{1} and i_{2}. Current i_{1} flows in the arm AB while i_{2} in arm AD. Current i_{1}, on reaching the junction B gets further divided into two parts (i_{1 }- i_{g}) and i_{g}, along branches BC and BD respectively. At junction D, currents i_{2} and i_{g} are added to give a current (i_{2} i_{g}), along branch DC. (i_{2} - i_{g}) and (i_{2} i_{g}) add up at junction C to give a current (i_{1} i_{2}) 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 K_{2}, deflection in the galvanometer becomes zero or current i_{g} 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 i_{1} flows in arms AB and BC and similarly same current i_{2} 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, i_{1}P + i_{g}G - i_{2}R = 0 ...(1)

Similarly, for the closed mesh BCDB, we get, (i_{1} - i_{g}) Q - (i_{2} + i_{g})S - i_{g}G = 0 ...(2) When bridge is balanced, i_{g} = 0. Hence eq. (1) & (2) reduce to

i_{1}P - i_{2}R = 0 or i_{1}P = i_{2}R ....(3)

i_{1}Q - i_{2}S = 0 or i_{1}Q = i_{2}S ...(4)

Dividing (3) by (4), we have, ....(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.**

**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

**Ex.23 Find the equivalent resistance between A and B**

**Sol. **This arrangement can be modified as shown in figure since it is balanced wheat stone bridge.

**8.1 Unbalanced Wheatstone Bridge**

**Ex**.**24**

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

Applying KCl at point B

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

Applying KCL at point E

⇒ 8y - 5x = 2v ...(2)

solving x & y ,

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

i = i_{3} i_{4} = =

Because, Therefore, R_{eq.} =

**Ladder Problem :**

Find the effective resistance between A & B ?

**Sol. **Let the effective resistance between A & B be R_{E} 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 R_{E}.

The equivalent network will be as shown below

The original infinite chain is equivalent to R in series with R & R_{E} in parallel.

R_{E}R + R_{E}^{2} = R^{2} + 2RR_{E} ⇒ R_{E}^{2} - RR_{E} - R^{2} = 0

**Ex.25**

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 kR_{E} (where R_{E} is the effective resistance)

Therefore, Effective R between A & B.

On solving we get

**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**

**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)

**II Method : folding symmetry**

**Therefore, **The potential difference in R between (B, C) and between (B, D) is same V_{C} = V_{D}

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

This is called folding.

Now , it is Balanced Wheatstone bridge

** **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**

**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.

Here V_{c} = V_{d}

hence C and D are same point

So, the circuit can be simplified as

Now it is balanced wheat stone bridge.

=

**Ex.28 Find the equivalent Resistance between A and B**

**Sol. **

Here V_{A} = V_{C} and V_{B} = V_{D}

Here the circuit can be simplified as

=

**Ans. **

** **

**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)

**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.

(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.

(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

**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.

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 :

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