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rva.raghav
H9P1: RESPONSE TO A DELAYED IMPULSE Note: In this problem we have chosen numbers for the part parameters to make it easier to compute an answer :-). By the way, it is also hard to arrange zero resistance, except with superconducting materials at very low temperatures. In the circuit shown below L=65.0H and C=6.24mF. The current source puts out an impulse of area A=2/π=0.64 Coulombs at time t=1.0s. At t=0 the state is: vC(0)=0.0 and iL(0)=1.0. The equation governing the evolution of the inductor current in this circuit is d2iL(t)dt2+1LCiL(t)=ALCδ(t−1.0) Can any one help me
callisto...can u help me
vision 89 can u help me
are u active vision89
please.......someone help
a)0.25 b)? c)? d)0 e)? f)0 g)0 h)0
0.250122093132 32.5 32.5 0 -102.05 0 0 0
In my case: H9P1 a - 0.25 b - 85 c - 85 d - 0 e - -267.03 f - 0 g - 0 h - 0
thankyou Abhay.....u are helping the best possible ways u can.......u helped me aerlier in multivibrators
but in case they are giving red mark only for b)32.5 c)32.5 e)102.05
even for b)85 c)85 e)267.03 giving red mark
Vision89, assuming: L=10H, f= 0.25hz, t= 9s vc= -15.7079632679
how to find b, c , & e plz tell the procedure with formula