In a Wheatstone’s bridge, three resistances P, Q and R are connected in three arms and the fourth arm is formed by two resistances S_1 and S_2 connect - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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In a Wheatstone’s bridge, three resistances P, Q and R are connected in three arms and the fourth arm is formed by two resistances S_{1} and S_{2} connected in parallel. The condition for the bridge to be balanced is:

Solution & Step-by-step Explanation

The balance condition for a Wheatstone bridge is \frac{P}{Q} = \frac{R}{S _{e q}} .Since S_{1} and S_{2} are in parallel, the equivalent resistance S_{e q} is: S_{e q} = \frac{S _{1} S _{2}}{S _{1} + S _{2}} Substituting this into the balance condition: \frac{P}{Q} = \frac{R}{\frac{S _{1} S _{2}}{S _{1} + S _{2}}} = \frac{R ( S _{1} + S _{2} )}{S _{1} S _{2}}

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In a Wheatstone’s bridge, three resistances P, Q and R are connected in three arms and the fourth arm is formed by two resistances S_{1} and S_{2} connected in parallel. The condition for the bridge to be balanced is:

\frac{P}{Q} = \frac{R}{S _{1} + S _{2}}

\frac{P}{Q} = \frac{2 R}{S _{1} + S _{2}}

\frac{P}{Q} = \frac{R ( S _{1} + S _{2} )}{S _{1} S _{2}}

\frac{P}{Q} = \frac{R ( S _{1} + S _{2} )}{2 S _{1} S _{2}}

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