Question

Two parallel resistors of 5ohm and 20 ohm are connected in left aem of the meter bridge. If the null point is 40 cm from left end, calculate the value of resistance connected in right arm

Answer 1

Answer:$R = 6\Omega$

Explanation:

Given that,

Length = 40 cm

$R_{1} = 5\Omega$

$R_{2} = 20 \Omega$

Now, the equivalent resistance is

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} +\dfrac{1}{R_{2}}$

$\dfrac{1}{R_{eq}} = \dfrac{1}{5\Omega} +\dfrac{1}{20\Omega}$

$R = 4 \Omega$

When bridge is balanced

$\dfrac{R_{eq}}{R} = \dfrac{l}{100-l}$

$\dfrac{4\Omega}{R} = \dfrac{40 cm}{100cm-40cm}$

$R = 6\Omega$

Hence,this is the required solution.