The resistances in two arms of the metre bridge are 5Ω and R Ω respectively, null point is obtained at length l 1 . When the resistance R is shunted with an equal resistance, the new balance point i

Question

The resistances in two arms of the metre bridge are 5Ω and R Ω respectively, null point is obtained at length l 1 ​ . When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l 1 ​ . The resistance R is

TestHub · Accepted Answer

For a metre bridge, the balancing condition is given by $\frac{R _{1}}{R _{2}} = \frac{l}{100 - l}$ .

Initially, the resistances in the two arms are $5Ω$ and $R$ .

$\frac{5}{R} = \frac{l _{1}}{100 - l _{1}}$ ... (1)

When $R$ is shunted with an equal resistance $R$ , the equivalent resistance in the right gap becomes $\frac{R \times R}{R + R} = \frac{R}{2}$ .

The new balancing length is given as $1.6 l_{1}$ .

$\frac{5}{\frac{R}{2}} = \frac{1.6 l _{1}}{100 - 1.6 l _{1}}$

$\frac{10}{R} = \frac{1.6 l _{1}}{100 - 1.6 l _{1}}$ ... (2)

Dividing equation (2) by equation (1), we get:

$2 = \frac{1.6 ( 100 - l _{1} )}{100 - 1.6 l _{1}}$

$2 (100 - 1.6 l_{1}) = 1.6 (100 - l_{1})$

$200 - 3.2 l_{1} = 160 - 1.6 l_{1}$

$1.6 l_{1} = 40$

$l_{1} = 25$ cm

Substituting the value of $l_{1}$ in equation (1):

$\frac{5}{R} = \frac{25}{100 - 25}$

$\frac{5}{R} = \frac{25}{75}$

$\frac{5}{R} = \frac{1}{3}$

$R = 15Ω$

Answer: $15$ $Ω$

Physics - Current Electricity Question with Solution | TestHub

Options:

Doubts & Discussion