Hi Michelle,
I'm wondering how the video was presented and why you arrived at somewhat complicated expression for computing the capacitance in a series RC circuit.
I haven't used the application version of Multisim for quite a long time and I only had the versions packaged with books. What I want to convey is I am not an expert who can provide precise solution to your problem, I just want to participate in this forum to share and learn some ideas.
First, I cannot see V and vout1 in the list of variables (2nd attached image). Maybe you've taken the expression from the video but you haven't renamed (or aliased) the voltage source and the node where the resistor and capacitor are connected. I think V and vout correspond to AC voltage source (V1) and capacitor voltage respectively, from Multisim's internal names these are V(1) and V(2). Also, I (common current) can be any of Multisim's internal names I(V1), I(R1), I(C1).
If we have the capacitor voltage and common current as phasors, the capacitance can be computed as
C = 1 / ( 2*pi * f * abs( V(2)/I(C1) ) )
In this context imag() will produce a result which is equal in magnitude to that generated by abs() but we prefer to not bother about the sign anymore.
If we have the capacitor voltage and common current as scalar measurements, the capacitance can be computed as
C = I(C1) / ( 2*pi * f * V(2) )
If we use the known quantities V(1) and R1, we can compute the capacitance as
C = ( V(1)-V(2) ) / ( 2*pi * f * R * V(2) )
In this case we only need (one) voltmeter / voltage probe.
Best regards,
G. Goodwin
