A) 6.28 A
B) 3.4 A
C) \[3.75\times {{10}^{-4}}A\]
D) 12.56 A
Correct Answer: A
Solution :
Displacement current \[{{I}_{d}}=\frac{dq}{dt}\] \[q={{q}_{0}}\sin \omega t;\] \[{{I}_{d}}=\frac{dq}{dt}={{q}_{0}}\omega \cos \omega t\] \[{{({{I}_{d}})}_{\max }}={{10}^{-6}}\times 2\pi \times {{10}^{6}}A=6.28A\]
You need to login to perform this action.
You will be redirected in
3 sec
