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NEETPhysicsCurrent Electricity

A Wheatstone bridge has three of its arms with a resistance of 10 each. The fourth arm needs to have an equivalent resistance of 10 to balance the bridge. List-I describes the current components present in the fourth arm, while List-II gives the modification required to achieve the balance condition. List-I (Current components in fourth arm) List-II (Modification required) (A) An ideal diode (forward-biased) in serie

Options

  1. A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  2. B(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  3. C(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  4. D(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct answer

D. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Step-by-step solution

For the Wheatstone bridge to be balanced with three arms of 10 each, the equivalent resistance of the fourth arm must also be 10 . (A) An ideal forward-biased diode has zero resistance. The arm's resistance is 0 + 20 = 20 . To reduce this to 10 , a 20 resistor must be connected in parallel ( 20 20 20 + 20 = 10 ). So, (A) matches with (III). (B) An ideal reverse-biased diode offers infinite resistance (open circuit). To make the arm's resistance 10 , a 10 resistor must be connected in parallel. So, (B) matches with

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