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(a) Obtain the condition under which the...

(a) Obtain the condition under which the current flowing in the current detecing device used in the circuit shown in figure, becomes zero.
(b) Describe briefly the device, based on the above condition. Draw a circuit diagram for this device and discuss, in brief, how it is used for finding as unknown resistance.

Text Solution

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(a) Applying Kichhoff's second law of the loop ABDA, we get,
Again applying Kirchhoff's second law to the loop BCDB, we get
` (I_1-I_delta)Q-(I_2_+I_delta)S-GI_delta=0`
In the balanced condition of the bridge,`I_delta=0`. The above equations become
` I_1P-I_2R=0 or I_1P=1_2R......(i)`
and ` I_1q-i_2S=0 or I_1Q=I_2S`
On dividing equation (i) by equation (ii), we get
` (P)/(Q)=(R)/(S)`
This proves the condition for the balanced Wheatstone bridge.
(i=b) Metre bridge or slide wire bridge. it is the simplest practical application of the Wheatstone bridge that is used to measure an unknown resistance.
Principal : Its working is based on the principle of Wheatsotne bridge. When the bridge is balanced,
` (P)/(Q)=(R)/(S)`
Construction : It consists of usually one metre long magnanin wire of uniform cross - section, stretched along a matre scale ficed over a wooden board and with its two ends soldered to two L-shaped thick copper strips A and C. Between these two copper strips, another copper strip is fixed so as to provide two gaps ab and the `a_1 b _1 ` . A resistance box R.B. is connected in the gap ab and the unknown resistance S is connected in the gap ` a_1 b _1 `.A source of emf ` epsi` is connected across AC. A movable jockey and a galvanometer are connected across BD, as shown in figure.
After taking out a suitable resistance R from the resistance box, the jockey is moved along the wire AC till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge. If P and Q are the resistances of the parts AB and BC of the wire, then for the balanced condition of the bridge , we have
` (P)/(Q) =(R)/(S)`.
Let total length of wire `AC=100 cm and AB= I cm`, Then`BC =(100-l) cm`. Since, the bridge wire is of uniform cross- section, therefore,
resistance of wire `prop` length of wire
` (P)/(Q)= ("resistance of AB")/("resistance of BC")`
` =(sigmal)/(singma( 100-l))=(l)/(100-l)`.
whee `sigma` is the resistance per unit length of the wire.
Hence, `(R)/(S)=(l)/(100-l)`
or ` S=(R(100-l))/(l)`

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