AT Any Time Padhai Academy JEE Mains Special Potentiometer & Meter Bridge
(c) To find the internal resistance of cell l arrangement 2nd arrangement R(known) by first arrangement , x11 by second arrangement IR = x12 =x(2 , also I =- E' xl r+R r+R R 21-l r+R R 2
The internal resistance of a cell is determined by using a potentiometer. In an experiment, an external resistance of 60S2 is used across the given cell. When the key is closed, the balance length on the potentiometer decreases from 72 cm to 60 cm. calculate the internal resistance of the cell Example K, K, SolutionA According to equation -x10 R+r From equation (i), (ii) and (i) we get here 10 72 cm ; 1 60 cm ; R 60 r-(60)|_-11 or r-12
METRE BRIDGE (USE TO MEASURE UNKNOWN RESISTANCE) If ABcm, then BC (100 ) cm Resistance of the wire between A and B Rcl [ Specific resistance and cross-sectional area A are same for whole of the wire or R= 1 where is resistance per cm of wire. Resistance Box (known) x (unknown) R.B thick connecting cdwires (strips) of negligible resistance (100 ()cm
Similarly, if Q is resistance of the wire between B and C, then Q 100 Q = (100-1) Dividing (1) by (2100 Applying the condition for balanced Wheatstone bridge, we get x-Ru or X= 100-1 Since R and are known, therefore, the value of X can be calculated.
Example In a meter bridge experiment, the value of unknown resistance is 2 . To get the balancing point at 40cm distance from the same end, the resistance in the resistance box will be (A) 0.5 (B) 3 (C) 20 (D) 80 Solution Apply condition for balance wheat stone bridge, P 100-40 P 100 Q 100- 2 40 Ans. : P 3C2.
In the given circuit, no current is passing through the galvanometer. If the cross-sectional diameter of the wire AB is doubled, then for null point of galvanometer, the value of AC would be [IT-JEE(Scr.) - 2003, 3/84] (A) 2X (B*) X (D) None 2
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