NEETPhysicsAlternating Current
An alternating current is given by the equation I = I₀ (120 t) . Match the instantaneous values of the current in List-I with the shortest time taken to reach them starting from t = 0 in List-II. List-I List-II (A) Peak value (i) 1 120 s (B) RMS value (ii) 1 240 s (C) Half of peak value (iii) 1 480 s (D) First return to zero (iv) 1 720 s Choose the correct answer from the options given below:
Options
- A(A)-(ii), (B)-(iv), (C)-(iii), (D)-(i)
- B(A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)
- C(A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
- D(A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)
Correct answer
D. (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)
Step-by-step solution
The given current equation is I = I₀ (120 t) . (A) Peak value ( I = I₀ ): I₀ (120 t) = I₀ (120 t) = 1 120 t = 2 t = 1 240 s. (B) RMS value ( I = I₀ 2 ): I₀ (120 t) = I₀ 2 (120 t) = 1 2 120 t = 4 t = 1 480 s. (C) Half of peak value ( I = I₀ 2 ): I₀ (120 t) = I₀ 2 (120 t) = 1 2 120 t = 6 t = 1 720 s. (D) First return to zero ( I = 0 after t=0 ): I₀ (120 t) = 0 (120 t) = 0 120 t = t = 1 120 s. Therefore, the correct matching is (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i). Answer: (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)