Question

ResistancesP Q R and S of values 2,3,2,4 form a wheat stone bridge,the resistance to be connected to S to have the bridge balanced is ??!

Answer 1

P / Q = R / S'
  S' = QR / P = 2*3/2 = 3
 
  1/S + 1/r = 1/S'
    1/4 + 1/r  = 1/3
     r = 12 ohms.
 connect a resistance of 12 ohms, in parallel with the resistance S (4 ohms) to balance the wheat-stone bridge.

Answer 2

Answer:

The resistance to be connected to S to have the bridge balanced is $12 \Omega$.

Explanation:

Given:

Resistance P = $2\Omega$

Resistance Q = $3 \Omega$,

Resistance R = $2 \Omega$,

Resistance S = $4 \Omega$.

Let the resistance is connected in parallel to the resistance S  be 'r'

Let the equivalent resistance of parallelly connected S and r be S'.

So, condition for a balanced Wheatstone bridge :

$\frac{P}{Q}=\frac{R}{S'}$$\implies S' =\frac{2*3}{2}=3\Omega$

$\implies P*S'=R*Q$

$\implies S' =\frac{R*Q}{P}$

The equivalent resistance of parallelly connected S and r :

$\frac{1}{S'} =\frac{1}{S} +\frac{1}{r}$

$\implies \frac{1}{3} =\frac{1}{4} +\frac{1}{r}$

$\implies \frac{1}{3} -\frac{1}{4} =\frac{1}{r}$

$\implies r=12\Omega$

So, the resistance to be connected to S to have the bridge balanced is $12 \Omega$.

#SPJ2