Home
Class 12
PHYSICS
In a Wheatstone bridge, three resistance...

In a Wheatstone bridge, three resistance, P, Q and R are connected in the three arms and fourth arm is formed by two resistance `s_(1) and s_(2)` connected in parallel. The condition for the bridge to be balanced for ratio `((p)/(Q))` is (given `R=10Omega, S_(1)=6Omega, S_(2)=3Omega`)

Text Solution

Verified by Experts

The correct Answer is:
5
If S is the equivalent resistance of `s_(1) and s_(2)` in parallel, then `S=(S_(1)S_(2))/(S_(1)+S_(2))`. For Wheat stone bridge to be balanced `(P)/(Q)=(R)/(S)or" "(P)/(Q)=(R)/(S_(1)S_(2)//(S_(1)+S_(2)))=(R(S_(1)+S_(2)))/(S_(1)S_(2))`
Promotional Banner

Similar Questions

Explore conceptually related problems

In a Wheatstone's bridge, three resistances P,Q and R 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

Three resistance P, Q, R each of 2 Omega and an unknown resistance S from the four amrs of a Wheatstone's bridge circuit. When a resistance of 6 Omega is connected in parallel to S the bridge gets balanced. What is the value of S ?

In a balanced wheatstone's network, the resistances in the arms Q and S are interchanged. As result of this:

When two resistances R_1 and R_2 are connected in series , they consume 12 W powers . When they are connected in parallel , they consume 50 W powers . What is the ratio of the powers of R_1 and R_2 ?

If three resistors of resistance 2Omega, 4Omega and 5 Omega are connected in parallel then the total resistance of the combination will be

In Wheatstone bridge, the resistances in four arms are 10Omega,10Omega,10Omega and 20Omega . To make the bridge balance, resistance connected across 20Omega is