Question

A wheatstone bridge is assembled with resistance in arm AB, BC and AD as 5 ohm, 15 ohm, 3 ohm respectively. what resistance should be inserted in arm CD to make it balanced????​

Answer 1

Given :

Resistance in arm AB = 5Ω

Resistance in arm BC = 15Ω

Resistance in arm AD = 3Ω

To Find :

What resistance should be inserted in arm CD to make it balanced!

Solution :

❖ In the balanced Wheatstone bridge, the resistance in arm BD is ineffective. The equivalent resistance of the balanced Wheatstone bridge between the point A and C is

$\dag\:\underline{\boxed{\bf{\gray{R_{eq}=\dfrac{(P+Q)(R+S)}{P+Q+R+S}}}}}$

For a balanced Wheatstone bridge;

$\dag\:\underline{\boxed{\bf{\pink{\dfrac{P}{Q}=\dfrac{R}{S}}}}}$

By substituting the given values;

$\sf:\implies\:\dfrac{P}{Q}=\dfrac{R}{S}$

$\sf:\implies\:\dfrac{5}{15}=\dfrac{3}{S}$

$\sf:\implies\:S=\dfrac{3\times 15}{5}$

$\sf:\implies\:S=3\times 3$

$:\implies\:\underline{\boxed{\bf{\green{S=9\:\Omega}}}}$