In the attached figure, "find the complex power transmitted to the load through each transformer." The solution (not shown) states that "with ΔV short-circuited, the current in each path is half the load current..."
If we are asked to measure the power transmitted to the load through each transformer, why do we need to do any short-circuiting whatsoever? Doesn't a short circuit imply that the power does NOT reach the load?
Source: Power Systems Analysis (Grainger/Stevenson/Chang)
In circuit analysis, short circuiting is one mental device to make this task easier.
It simplifies the circuit by replacing the shorted part with a wire. (A dual operation is isolating a suitable part of the circuit.)
In Figure 3.25, I interpret with ΔV short-circuited as without voltage source ΔV:
The part between \$+V_1\$ and \$+V_2\$ is just two j0.1 in parallel.
This way, establishing the complex power "transmitted to the load" (with \$+S\$ closed) should not be hard given \$+V_1\$ (by waveform, voltage, frequency).
In all likelihood, the solution (not shown) goes on to compute what changes when "reintroducing" ΔV.
How is a short circuit useful at all for measuring the power transmitted through a transformer?
By taking short circuit and open circuit measurements in a circuit, one can find impedances (including mutual impedances). Knowing voltages and impedances, one can find power.
Doesn't a short circuit imply that the power does NOT reach load?
Yes, a short circuit may have current through it, but not voltage across it. As a result, there is no power dissipated in an ideal short circuit. However, if one measures the current through the short circuit, one may use the found value to calculate the impedance of other parts of the circuit, and the impedance values thus found, may be useful in calculating the power in the circuit with a non-zero load instead of the short circuit.
