A) (1)
B) (2)
C) (3)
D) (4)
Correct Answer: C
Solution :
[c] \[{{i}_{1}}=\frac{dq}{dt}=\frac{d}{dt}(2\,\cos \,4t)=-8\,\sin \,4t\], \[{{i}_{2}}=\frac{dq}{dt}=\frac{d}{dt}(4\,\cos \,t)=-4\,\sin \,t\], \[{{i}_{3}}=\frac{dq}{dt}=\frac{d}{dt}(3\,\cos \,4t)=-12\,\sin \,4t\], \[{{i}_{4}}=\frac{dq}{dt}=\frac{d(4\,\cos \,2t)}{dt}=-\,8\,\sin \,2t\] Clearly, amplitude of \[{{i}_{3}}\] is greatest.
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