Four resistors of resistance `15 Omega, 12 Omega, 4 Omega and 10 Omega` arc connected in cyclic order to form a Wheatstone bridge. The resistance (in Q) that should be connected in parallel across the `10 Omega` resistor to balance the Wheatstone bridge is

Options

A. `10 Omega`
B. `5 Omega`
C. 15` Omega`
D. `20 Omega`

Answer

A

Explanation

`10 Omega`

suppose P = 15 `Omega, , Q = 12 Omega`
`R = 4 Omega , S = (10r)/(10 + r) `
Suppose, bridge achieves balanced condition by connecting resistor r in parallel to `10 Omega`
For balaned condition of Wheatstone bridge,
`therefore (P)/(Q)= (S)/(R)`
` (15)/(12) = (10r)/((10 + r) xx 4)`
`therefore (5)/(4) = (10r)/(40 + 4r)`
`therefore 5 = (10r)/(10 + r)`
`therefore 50 + 5r + 10 r`
`therefore 50 = 5r therefore r = 10 Omega`