Howdy Everyone!
Im having trouble with the following interesting question especially with part (c)
(a) For this part I realised the steady state principles and applied open and short circuits etc and used the voltage divider rule to get 2V and 1 A, is this correct?
(b) Redrawing this I got
6V turned into 6/s
4Ω stayed the same
1H turned into s
0.05F turned into 1/0.05s
Also for the initial conditions i added a current source under each inductor i(0-) and capacitor vc(0-).
(c) How do I find this Vout? I can see that its the voltage at the node connecting Ir,Il,Ic. So at this node Ir + Il + Ic = 0? From KVL/KCL?
How do I take into the account the -ve sign of the current source from the inductor.
Would it be Ir - Il + Ic = 0?
Ir = (Vo - 6/s)*1/4 ?????
Il = ?
Ic = ?
Then how would I get an expression for Vout? Hence the complete solution?
Cheers for the help in advance!
Im having trouble with the following interesting question especially with part (c)
(a) For this part I realised the steady state principles and applied open and short circuits etc and used the voltage divider rule to get 2V and 1 A, is this correct?
(b) Redrawing this I got
6V turned into 6/s
4Ω stayed the same
1H turned into s
0.05F turned into 1/0.05s
Also for the initial conditions i added a current source under each inductor i(0-) and capacitor vc(0-).
(c) How do I find this Vout? I can see that its the voltage at the node connecting Ir,Il,Ic. So at this node Ir + Il + Ic = 0? From KVL/KCL?
How do I take into the account the -ve sign of the current source from the inductor.
Would it be Ir - Il + Ic = 0?
Ir = (Vo - 6/s)*1/4 ?????
Il = ?
Ic = ?
Then how would I get an expression for Vout? Hence the complete solution?
Cheers for the help in advance!