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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 |
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Answer» `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 = 5rtherefore r = 10 Omega` |
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