# Changing the power factor of a circuit when connected to transformer

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.

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.

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?

• (1) What is "a series inductor ect."? Do you mean "a series inductor, etc. (et cetera)"? (2) Can you clarify in the last paragraph whether the inductance is added to the previous 3 Ω value or is that removed? – Transistor Sep 21 '17 at 1:15
• Okay I clarified it by editing. I meant adding a series inductor to the 3 Ohm for instance to adjust cosfi as 0.5. So the 3 Ohm will stay there but now we have an inductor as well which caused 0.5 cosfi. See my comment to the user Tony below. Do you agree? thanks – GNZ Sep 21 '17 at 10:01

## 2 Answers

This is a pictorial answer which may help your thinking. simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) 3 Ω resistor only. (b) 3 Ω resistor and 3 Ω inductance.

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 $\sqrt 2 R$.

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.

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.)

• I see but for cosfi to be 0.5 the inductor you chose should have been around 5.2Ω. Z then becomes 6Ω for 0.5 cosfi. Current driven from secondary becomes 48/6 = 8A. (?) – GNZ Sep 21 '17 at 15:53
• Sorry, I started working on this before your clarification edit and I wasn't too sure I understood the question. Your comment above looks correct. The principle I wanted to get across was that the current and VA would decrease as the load inductance increases. So, "Would the current increase?" No. It would decrease. – Transistor Sep 21 '17 at 17:28

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.

• Yes I have familiarity with phasors. So currently cosfi=1 so the P = 48*16 = 768W and since at the beginning cosfi = 1, S = 768VA. But when the load is made cosfi=0.5 will the P remain same and Q will appear? So the S required will be 2*768 = 1536VA. Yes then S increases but now it is exceeding 1000VA rating. Is my way of thinking correct? – GNZ Sep 20 '17 at 14:45