CBSE Class 12 · 2009 · Current ElectricityShort Answer

In a bridge, with two unknown resistances X and Y inserted, the null point D is

40 cm

from end A. When a

10 Ω

resistance is connected in series with X, the null point shifts by

10 cm

. Find the position of the null point when the

10 Ω

is instead connected in series with Y, and determine X and Y.

Solution

At the first null,

\frac{X}{Y} = \frac{40}{60} = \frac{2}{3}

, so

X = \frac{2}{3} Y

. Adding

10 Ω

in series with X (giving

X + 30

after accounting for the stated shift to

50 cm

\frac{X + 30}{Y} = \frac{50}{50} = 1 \Rightarrow X + 30 = Y

. Solving with

X = \frac{2}{3} Y

gives

Y = 90 Ω

and

X = 60 Ω

. With

10 Ω

in series with Y,

\frac{X}{Y + 30} = \frac{60}{120} = \frac{60}{100 + 20}

, i.e. the null point is at

60 cm

— shifted

20 cm

to the right.

CBSE Class 12 Board · 2009 · Subjective