Any help much appreciated here, I cannot understand the following formulas at all. If there is anyone who can please explain in very SIMPLE words I would be very grateful. My exam is next week and I am still stumped on the following.VP/VS NP/NS IS/IP ok this bit I understand, but then here are the QAs that don't make sense to me:Q1) A single phase transformer has 1250 primary turns and 200 secondary turns. Primary V I are 230V and 3A. What are the secondary V I?A) 1250/200 230/vs transpose too: VS 200 X 230/1250 36.8V Why has it been turned upside down?230/36.8 Is/3transpose too: Is 3 X 230/36.8 18.75A Why??The following Q I am even more stuck withQ2) Single phase transformer has a rating of 300/230V, 15KVA. What is the full load of primary and secondary?A)S VI Where did S come from?!Ip 1500/3300 4.55A KVA - is this taken as the V? Not I? Is that why Is 1500/230 65.22A it has been reversed? I'm very confused, please can someone help explain this process to me?
This seems to be about transposing equations rather than transformers! Let's go through it one step at a time. VP/VS NP/NS IS/IP OK So far. Q1) A single phase transformer has 1250 primary turns and 200 secondary turns. Primary V I are 230V and 3A. What are the secondary V I? Voltage first. So you are given Np 1250, Ns 200 and Vp 230 VP/VS NP/NS 230 / Vs 1250 / 200 Multiply both sides of this equation by Vs Vs * 230 / Vs Vs * 1250 / 200 Simplifies to: 230 Vs * 1250 / 200 Multiply both sides by 200 200 * 230 200 * Vs * 1250 /200 Simplifies to: 200 * 230 Vs * 1250 Divide both sides by 1250 200 * 230/1250 Vs * 1250/1250 Simplifies to: 200* 230/1250 Vs Vs 36.8 volts OK. Now the current. You are told that Ip 3 amps VP/VS NP/NS IS/IP Now you know Vs 36.8 volts, you have a choice as to how you set the equation up. You could use either: VP/VS IS/IP 230 / 36.8 Is/3 or NP/NS IS/IP 1250 / 200 Is/3 Either way, multiply both sides by 3 3 * 230 / 36.8 Is or 3 * 1250/200 Is Both give Is 18.75 amps Q2) Single phase transformer has a rating of 300/230V, 15KVA. What is the full load of primary and secondary? There is a typo in this question. The rating is really Vp/Vs {which is supposed to read 3300 V / 230 V} and the maximum value for the product of (Vp * Ip) 15000 (volts * amps) { K 1000 } {Which for our purposes is the same as (Vs * Is) } So we know that on the primary side: 3300 * Ip 15000 Divide both sides by 3300 Ip 15000 /3300 Ip 4.55 A and on the secondary side: 230 * Is 15000 Divide both sides by 230 Is 15000 / 230 Is 65.22 A Hope that helps.
With good intention, I would criticize UR copying of Q2 :} From the primary (Ip) and secondary (Is) currents listed, this transformer's Voltages should have been: 3300V/230V at 15 KVA note the V(volts) after EACH of the two voltages separated by / and that the primary voltage must be 3300 V and not 300 V in order to agree with Ip Is. I don't know where S comes from, but V?I represents the power rating of a given coil (primary or secondary) in a transformer. If the voltage ratio is 3300/230 14.3478/1 then the current ratio must be the INVERSE 1/14.3478 in order that the power on each side (primary and secondary) is the SAME and equal to 15,000 KVA. {this math assumes 100% efficiency which is an IDEAL case for a transformer}
