The potential difference across a 2H inductor as a function of time is shown in the figure. At t=0, current is zero. Choose the correct statement(s).
A
Current at t=2s is 5A
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Current at t=2s is 10A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Current versus time graph across the inductor will be
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Current versus time graph across the inductor will be
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct options are A Current at t=2s is 5A C Current versus time graph across the inductor will be
We know that potential across inductor, V=LdIdt(1) ⇒∫dI=∫20VdtL Now ∫20Vdt= Area under the V-t graph for t=0−2sec Area, A=12×2×10=10 Therefore ⇒∫dI=I=AL=102=5A Hence option A is correct. For current variation lets consider voltage variation in 2 cases i.e 0-2 sec and 2-4 sec. Case 1: For t=0−2 sec Equation of straight line passing through origin, V=mt+C, where m is slope and C is y-intercept From graph, m=tanθ=102=5 and C=0 Therefore, V=5t Substituting V in equation (1) and integrating, ∫dI=∫VdtL I=1L∫5tdt=5t22L Its evident that the variation is not linear and hence option D is incorrect. Also the curve should show increasing slope from 0-2 sec. Case 2: For t=2−4 sec Variation of V with t can be given using staright line equation as, V=−5t+20 ∫dI=∫VdtL I=1L∫(−5t+20)dt=1L(−5t22+20t) The above equation shows decreasing slope and so does the graph in option C. Hence option C is correct.