I think the OP has the right numerical answer (although I'm having trouble following how he got it.) I did the simple voltage divider approach, but since it should not matter from which direction the voltage division is done, I did it both ways (Eqns 2 & 4) to ensure the answer was the same both...
But suppose that there was no voltmeter connected; yet the problem asked you to find the unknown Z such that nodes D & F were at the same potential. So what does the presence of the voltmeter actually imply?
First consider how one might simplify the problem. The voltmeter has infinite impedance so C4 is not relevant. The voltmeter reads zero and a zero signal has no phase, so the phase of the current source is not relevant. But the current source does cause a voltage drop from Point-A to Point-B...
Of course, you would. But you can only make that statement after having done the mesh analysis and solved the circuit equations. One might find that the mesh current flows in the opposite direction of the real current.
This question suggests an unfamiliarity with the method of mesh current analysis. Suffice it to say that mesh currents are not real currents but merely an analysis tool for deriving the equations that govern a circuit's operation. The other prime method for deriving circuit equations is node...
But not all possible gate functions are described in that link, and only two of those described are universal. Note that to be a universal gate function the gate does not necessarily need to be commercially available nor even especially useful - just that it can (theoretically) be used to...
So what is excluded? Any 2-input gate has 4 possible input combinations and 2 possible outputs for each input combination. But the gate function is determined by selecting just one of the possible outputs for each of the 4 possible input combinations; therefore, the gate's function can be...
On first seeing this circuit, my first inclination was to perform a nodal analysis in the complex frequency domain, because that is how I always analyze circuits. But realizing that this particular problem is in the steady-state frequency domain, it can be redrawn as simple resistors &...
True; however, they do teach DeMorgan's Theorem for a reason. But one thing I was never taught is the use of the Boolean minus "-" operation. There is the AND gate, the OR gate, the NOT gate. Where does one get a MINUS gate?
The original question was how did the author arrive at the circled expression? It was done by multiple application of DeMorgan's Theorem, {When breaking an overbar, the operation enclosed by the overbar changes between AND / OR}.