Question

With a resistance of 'X' in the left gap and a resistance of 9 Ω in the right gap of a meter bridge, the balance point is obtained at 40 cm from the left end.
In what way and to which resistance 3 Ω resistance be connected to obtain the balance at 50 cm from the left end?

Options

• In series with 9 Ω

• Parallel to X Ω

• In series with X Ω

• Parallel to 9 Ω

Answer 1

In series with X Ω

Explanation:

The balanced condition of bridge is shown below in the first case

Then, `"X"/9 = 40/60 => "X" = 2/3 xx 9 = 6 Omega`

To get the balance point at 50 cm from left end, the balanced condition becomes,

`("X"')/9 = 50/50 = 1`

= X' = 9 Ω

As, in series combination resistances are added, so to get 9 Ω at left gap the 3 Ω resistance should be added in series with X Ω (6 + 3 = 9 Ω).