Changing the power factor of a circuit when connected to transformer - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-09-20T21:04:03Z https://electronics.stackexchange.com/feeds/question/330287 https://creativecommons.org/licenses/by-sa/4.0/rdf https://electronics.stackexchange.com/q/330287 0 Changing the power factor of a circuit when connected to transformer Genzo https://electronics.stackexchange.com/users/161776 2017-09-20T14:26:30Z 2017-09-21T13:13:11Z <p>There is a step-down transformer with a primary voltage rating of 120 volts, a secondary voltage rating of 48 volts, and a VA rating of 1 kVA (1000 VA). The maximum secondary winding currents can be determined as such: 1000VA / 48V = 20.83A.</p> <p>So if I connect a pure resisitive circuit/load to the secondary which is 3 Ohm, the current through the secondary winding will be 48/3 = 16A rms which is smaller than 20.83A rms. I think it means this situation is acceptable.</p> <p>Now when this 3 Ohm is connected to the secondary and if I play with the circuit adding a series inductor ect.(keeping 3 Ohm and without adding any extra resistor) and make the load's power factor 0.5 instead of 1(pure resistive) what would happen? Would the current increase? How can this situation be analyzed?</p> https://electronics.stackexchange.com/questions/330287/-/330289#330289 0 Answer by Sunnyskyguy EE75 for Changing the power factor of a circuit when connected to transformer Sunnyskyguy EE75 https://electronics.stackexchange.com/users/17574 2017-09-20T14:39:15Z 2017-09-20T14:39:15Z <p>This increases the apparent power by phaser increase in length, so that VA is increased thus you must derate the supply for real power. I assume you know how to do phasor geometry.</p> https://electronics.stackexchange.com/questions/330287/-/330511#330511 1 Answer by Transistor for Changing the power factor of a circuit when connected to transformer Transistor https://electronics.stackexchange.com/users/73158 2017-09-21T13:13:11Z 2017-09-21T13:13:11Z <p>This is a pictorial answer which may help your thinking.</p> <p><img src="https://i.stack.imgur.com/ePd0C.png" alt="schematic"></p> <p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fePd0C.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p> <p><em>Figure 1. (a) 3 Ω resistor only. (b) 3 Ω resistor and 3 Ω inductance.</em></p> <p>In Figure 1b we can see that the impedance of the inductor is at right angles to that of the resistor. Since the impedances are equal in magnitude we can show by Pythagoras Theorem that the combined impedance of the two elements has a magnitude of \$\sqrt 2 R \$.</p> <p>From \$I = \frac {V}{Z} = \frac {V}{\sqrt{2}R} \$ we can see that the current is \$\frac {1}{\sqrt{2}} \$ (about 70%) of that with the resistor on its own.</p> <p>So, for your 48 V supply into 3 Ω we would expect 16 A as you calculated. For 3 - j3 Ω we would expect \$I = \frac {16}{\sqrt{2}} = 11.3 \; A \$. (Someone correct me if I have the sign wrong.)</p>