Question

Four resistances 6Ω,6 Ω,6 Ω and 18 Ω form a Wheatstone’s meter bridge. Find

the resistance which should be connected across the 18 Ω resistance will balance

the network.​

Answer 1

Answer:

9Ω

Explanation:

Let the that should be R. Effective resistance of 18Ω and R is =

= 18R/(18 + R)

For the system to be form a Wheatstone’s meter bridge:

⇒ 6/6 = 6/[ 18R/(18 + R) ]

⇒ 1 = 6(18 + R)/18R

⇒ 1 = (18 + R)/3R

⇒ 3R = 18 + R

⇒ 2R = 18

⇒ R = 9

  The required resistance should be 9Ω

Answer 2

Given :-

Four resistances 6Ω,6 Ω,6 Ω and 18 Ω form a Wheatstone’s meter bridge

To Find :-

The  resistance which should be connected across the 18 Ω resistance will balance  the network.​

Solution :-

Let the resistance be r

$\sf \dfrac{6}{6} = \dfrac{6}{ 18r/(18 + r) }$

$\sf \dfrac{6}{6} = \dfrac{6}{18r} \times\dfrac{1}{18 +r}$

$\sf 1 = \dfrac{6(18 + r)}{18r}$

$\sf 1 = \dfrac{108 + 6r}{18r}$

$\sf 1= \dfrac{108 + r}{3r}$

$\sf 3r = 108 + r$

$\sf 3r - r = 108$

$\sf 2r = 108$

r = 108/2

r = 54 ohm