In a meter Bridge experiment resistance x is connected in the right gap. When R 1 and R 2 are connected in the left gap separately the balance points are 40 c m and 50 c m respectively from the left end. Now if both R 1 & R 2 are connected in series in the left gap with x in the right gap then what will be the new balance point from the left end (in cm )?

Step-by-step solution

R 1 x =   40 60   =   2 3 . . . . 1 ,   R 2 x =   50 50   =   1   . . . . .   2 R 1 + R 2 x =   l 100 – l .... (3) Add eq (1) & (2) R 1 + R 2 x = 2 3   +   1   =   5 3 ... (4) Placing eq (4) in (3) 5 / 3   = l 100 – l ⇒   500 – 5 l   =   3 l ⇒   8 l = 500 ∴ l =   62.5   c m