At 11:51, why do you take the voltage over the 16ohm capacitor as the voltage between a and b? Wouldn't it not be the same because of the 10 ohm capacitor? I understand the open circuit is caused by the current source being set to 0 but why does that 10 ohm capacitor not interfere with the voltage?
Because when that constant current source drives current through the Thevenin equivalent resistance, the current i going from point b to point a, but we want to know the voltage from point a to point b.
At 11:51, why do you take the voltage over the 16ohm capacitor as the voltage between a and b? Wouldn't it not be the same because of the 10 ohm capacitor? I understand the open circuit is caused by the current source being set to 0 but why does that 10 ohm capacitor not interfere with the voltage?
There will be no current through the 10ohm capacitor, so no voltage across it.
At 10:30 why is the current from the source taken as negative?
Because when that constant current source drives current through the Thevenin equivalent resistance, the current i going from point b to point a, but we want to know the voltage from point a to point b.
8:40 u took capacitor as 10 ohms while calculating the th impedance shouldnt it be -j10 ohms?
You're right I messed that up, I meant for it to be a resistor. At 7:20, I called it a resistor but drew it as a capacitor. Sorry about that.
You presumed the radian frequency of 1 radians per seconds which means your f= 1/2pi cycles per seconds from w=2pi.f
8:40 gave me (14.769 - 35.846j), or 29.768 angle of minus 60.255
then plus 10 becomes 24.769 -35.846j
thanks :-)