Thevenin Equivalent Network Theorm is a powerful tool. Here are two examples to show how to find analytical solutions of circuits which seem difficult.
Here we have an AC voltage source with internal resistance connected to a resistor (the load) through a linear transformer. Transformer parameters are given as self and mutual inductances.
The secondary node voltage can be found by nodal analysis, which is rather messy.
However, if we use Thevenin equivalent network theorem, the analytical solution in phasor domain can be found in a simple and elegant way.
There are two steps to find the Thevenin equivalent network of the simulation circuit on the left side of the green line:
Once $Z_{EQ}$ and $E_{OC}$ are found, the Thevenin equivalent network (the part on the left side of the green line in Figure 3) can be used to find the load voltage $V_L$.
It can be observed from the plot that the positve and negative peaks of $V_L$ are slightly different since the system has not yet reached steady state at $t=2$ seconds. If you run the simulation for a longer period of time (such as for 5 to 6 seconds and plot the last 0.1 seconds) those peaks will almost be the same.
It can also be seen that the peak values of $V_L$ from the analytical solution and the simulation are very close.
If the time step is reduced by half to 25 μS, the simulation error only reduces slightly, which is an indication that the simulation time step of 50 μS is already sufficiently small.
This approach could be use to evaluate the accuracies of the simulation:
If the simulation results from the above two runs are close enough for you, you can rest assured that the time step Δt is sufficiently small.
Wheatstone bridge is used to measure resistance.
Three of the four resistors have known and precise values, the other's resistance value is to be determined.
A current meter (with a serial resistor to limit the current) is connected between node A and node B. When current $I_{AB}$ is zero the bridge is balanced.
The resistance values of those four resistors have to meet certain condition for the bridge to be balanced. The easiest way to find the balance condition is to use Thevenin equivalent network theorem:
If $E_{OC}$ between node A and B is zero, current $I_{AB}$ will always be zero.
There are two steps to find the Thevenin equivalent network of the Wheatstone bridge with the $R_L$ as the load:
An ideal current is used to derive the balance condition for simplicity reasons. If a voltage source with internal resistance is used instead, it can be converted into an ideal current in parallel with a resistor ($R_S$) (Norton equivalent). The existance of $R_S$ will not have any effects on the balance condition. It would, however, make it much more difficult to find $R_{EQ}$.
If an AC source is used the bridge can be configured to measure inductance or capacitance values.
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