For the RC circuit shown on the picture, calculate the U_{2},
knowing the sinusoidal change of U_{1}.

Solution:

- Concepts:

AC circuits - Reasoning:

The circuit shown is a simple, two terminal, AC circuit. - Details of the calculation:

The impedance of the capacitor is Z_{C}= -i/(ωC) = -iX and the impedance of the resistor is Z_{R}= R.

In each branch of the circuit (left and right) I = U_{1}/(Z_{C}+ Z_{R}).

U_{2}= IZ_{R}- IZ_{C}= U_{1}(Z_{R}- Z_{C})/(Z_{R}+ Z_{C}) = U_{1}(R + iX)/(R - iX) = U_{1}e^{iφ}.

tan(φ/2) = X/R = 1/(ωRC).

U_{2}= Re(U_{1}exp(iφ)).

U_{2}= U_{0}cos(ωt + φ).

The voltage amplitude stays the same, the phase is shifted.

In electronic circuits it is often desired to provide a sinusoidal voltage of
constant amplitude but variable phase. A circuit which accomplishes this is
called a phase shifting network. One example of such a network is shown in the
figure. Show that the voltage measured between terminals A and B
has half the amplitude of the input voltage, and a phase which may be adjusted
between 0^{o} and 180^{o} by changing the resistance R'.

Solution:

- Concepts:

AC circuits - Reasoning:

We have a two-terminal network and a AC generator, generating a sinusoidal voltage. - Details of the calculation:

V_{in}= I_{1}Z_{1}= I_{2}Z_{2}. Z_{1}and Z_{2}are in parallel.

Z_{1}= 2R, Z_{2}= R' + 1/(iωC) = R' - i/(ωC).

V_{in}= V_{0}exp(iωt), I_{1}= (V_{0}/2R)exp(iωt), I_{2}= (V_{0}/(R' - i/(ωC)))exp(iωt).

V_{A}= V_{in}- I_{1}R, V_{B}= V_{in}- I_{2}R'.

V_{out}= V_{A}- V_{B}= -I_{1}R + I_{2}R'.

V_{out}= [V_{0}R'/(R' - i/(ωC)) - V_{0}/2]exp(iωt)

= [(V_{0}R' - V_{0}R'/2 + iV_{0}/(2ωC))/(R' - i/(ωC))]exp(iωt).

V_{out}= (V_{0}/2)(R' + i/(ωC))/(R' - i/(ωC))exp(iωt) = (V_{0}/2)exp(i(ωt + 2φ)).

tanφ = 1/(R'ωC

A student measures voltage drops in an AC circuit using three voltmeters and
arrives at the following measurements:

Upon viewing these measurements, the student becomes very perplexed. Aren't voltage drops supposed to add in series, just as in DC circuits? Why, then, is the total voltage in this circuit only 10.8 volts and not 15.74 volts? Explain the reason for the "discrepancy" in the voltage measurements. Show that at any point in time Kirchhoff's loop rule is satisfied.

Solution:

- Concepts:

AC circuits - Reasoning:

We have a series RC circuit. - Details of the calculation:

V = V_{0}exp(iωt), I = V/(Z_{C}+ R).

V_{R}= IR = VR/(Z_{C}+ R), V_{C}= IZ_{C}= VZ_{c}/(Z_{C}+ R), V_{R}+ V_{C}= V.

The AC voltages add in series, but phase must be considered. Multi-meters display no phase information, just RMS voltages.

Z = R - i/(ωC) = (R^{2}+ 1/(ωC)^{½})exp(iφ) = |Z|exp(iφ). Here tanφ = -1/(ωRC)

V_{R}= VRexp(-iφ) /|Z|, V_{C}= V(-i/(ωC))exp(-iφ)/|Z|,

V_{R}+ V_{c}= V|Z|exp(iφ)exp(-iφ)/|Z| = V.