In a Wheatstone’s bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be

Question

In a Wheatstone’s bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be (A)  (P/Q) = (2R/S1 +S2)  (B)  (P/Q) =(R(S1 +S2)/S1 S2)  (C)  (P/Q) =(R(S1 +S2)/2S1 S2)  (D)  (P/Q) = (R/S1 +S2)

Tardigrade Admin · Accepted Answer

For balanced Wheatstone's bridge (P/Q)=(R/S) here S=S1 | S2=(S1 S2/S1+S2) ⇒ (P/Q)=(R(S1+S2)/S1 S2)

Q. In a Wheatstone’s bridge, three resistances $P, Q$ and $R$ are connected in the 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 will be

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

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

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

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

For balanced Wheatstone's bridge
$\frac{P}{Q} = \frac{R}{S}$ here $S = S_{1} ∥ S_{2} = \frac{S _{1} S _{2}}{S _{1} + S _{2}}$
$\Rightarrow \frac{P}{Q} = \frac{R ( S _{1} + S _{2} )}{S _{1} S _{2}}$

Q. In a Wheatstone’s bridge, three resistances P,Q and R are connected in the three arms and the fourth arm is formed by two resistances S1​ and S2​ connected in parallel. The condition for the bridge to be balanced will be

QP​=S1​+S2​2R​

QP​=S1​S2​R(S1​+S2​)​

QP​=2S1​S2​R(S1​+S2​)​

QP​=S1​+S2​R​