Alright thanks for the help and clarification :)
Nope, no idea at all on these, going to review my notes and lectures and see if there's anything I've missed
So am I right to assume since the signals are "shorted to ground", that after current passes through Rs and the first resistor, it is "curtailed" by any inductors and no other resistors or the load receives any current, which is why I put zero.
In regards to C and D, don't think I've learned...
Homework Statement
Homework Equations
At low voltage frequency (inductors) L become short circuit and (capacitors) C becomes open circuit
At high voltage frequency (inductors) L become open circuit and (capacitors) C becomes short circuit
The Attempt at a Solution
Part A
At low frequency...
@gneill also do we now take into account the current source and resistor on the left? since there is no longer a short circuit, as there will now be a current source there
So can it simply be a source transformation from your thevenins? with the new value for I being I=V/Z with Z being from the previous answer? and V being from the voltage source
Alright thanks,
And for nortons, is it right to assume that all the current in the circuit runs into the load:
since you have to short circuit the load to find the current for all the circuit's impedance
and then divide the voltage provided by this impedance to find the circuits current,
and...
The question says (Use peak values for the voltages and currents in your calculation)
So shouldn't you multiply 70/71 by the peak value (and not the phasor which was done) for V which is 10? and leave it as no real part? so 0 + j( 70/71 x 10 )
or why else does it clearly say use the peak values?
[moderator note: partial solution removed --- Please do not solve the Original Posters problems for them. Hints and suggestions are fine, but they must do the work themselves.]
Does the fact the impedances only have an imaginary part make a difference when working out VT using the voltage...