Four resistances of 15Ω, 12Ω, 4Ω and 10Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10Ω to balance the network is _____Ω.

Solution

Physics sir · 4 steps

In cyclic order, opposite arms multiply to the same value. Make the 10 Ω arm work out to 5 Ω.

1. Label the arms. Going round the network, P = 15 Ω, Q = 12 Ω, R = 4 Ω, S = 10 Ω. P and R are opposite arms, and so are Q and S.

   P/Q = S/R ⇒ P × R = Q × S

2. Check the balance. P × R = 15 × 4 = 60, but Q × S = 12 × 10 = 120. For balance we need 12 × S' = 60, so the 10 Ω arm must become S' = 5 Ω.
3. Find the shunt. With X in parallel with 10 Ω, 10X/(10 + X) = 5. This gives 10X = 50 + 5X, so X = 10 Ω.
4. So. A resistance of 10 Ω in parallel with the 10 Ω arm balances the network. Two equal resistors in parallel halve the value.

The common slip: 4.7. Using P/Q = R/S for adjacent arms gives a required arm of 3.2 Ω and a shunt of about 4.7 Ω instead of 10 Ω.

The idea to keep. For a Wheatstone network in cyclic order, the balance check is quick: the products of opposite arms must be equal. A parallel resistor can only lower an arm, so fix the arm that is too large.

Answer from the NTA final answer key. Solution by PrepWiser.