Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at 20 cm. When a resistance of 15 is connected in series with the smaller resistance of the two, the b…

Question

Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at 20 cm. When a resistance of 15 is connected in series with the smaller resistance of the two, the balancing point shifts to 40 cm. The value of smaller resistance is

Accepted Answer

Correct answer: A. (a) L=20 cm ;L^ '=40 cm Using R_1R_2= L100-L= 20100-20= 2080 ⇒R_2=4R_1 R_1^ 'R_2^ '= L^ '100-L^ '= 40100-40= 4060= 23 Here R_1 ^ '=R_1+15 and R_2 ^ '=R_2=4R_1 R_1+154R_1= 23 3R_1+45=8R_1 ⇒R_1=9 Smaller resistance

$Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at 20 cm . When a resistance of 15Ω is connected in series with the smaller resistance of the two, the balancing point shifts to 40 cm . The value of smaller resistance is$

Solution

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