Question one's electrical

The 240V output of a transformer secondary winding is fed in to a 25Ω load Step one- find the important information from above-  ·         We know that 240v is the voltage and 25Ω so these are the first two bits of data we will put in our key (see below)            Then we have to use our formula to find the current  Voltage- 240V Resistance- 25Ω                       V Current= 9.6A                                                  voltage = I x R 240/25= 9.6                             I x R                 Current = V/R                                                                          Resistance = V/I   Using this formula we have worked out that the current is 9.6A ( see in green above) Now you can work out the questions:a) Calculate the power in the loadb) if the primary voltage is 400V calculate the current  

a)    Calculate the power in the load.Power = V x IP = 240 x 9.6= 2,304 WP= 2304Wb)    If the primary voltage is 400V calculate the primary currentStep one-·         From the question above we know that to find the Power we have to multiply the voltage by the current which gives us 2304W·         We can now use this information to find the primary current. We have to rearrange the formula.          We know that if you divide 2304W by the secondary voltage of 240v it gives us the secondary current of 9.6A (all information that            you have been provided with)        We know that the primary voltage is 400V so in order to find the primary current we divide the power (2304W) by the primary           Voltage (400v)              2304/400= 5.785A              5.76A is the current 

 Question1 The 300V output of a transformer secondary winding is fed in to a 35Ω load-   find the important information, put it in a key and find the missing information before answering the question Voltage- 300V Resistance- 35Ω                        Current= 8.6A                                                  voltage = I x R 300/35= 8.6                                                     Current = V/R                                                                          Resistance = V/I Now we know the current we can go on to answer the questions:a) calculate the power in the loadb) if the primary Voltage is 400V calculate the primary currentc)    The secondary is wound with 340 turns, how many turns are there on the primary. d) The primary voltage falls to 391 volts. How many turns need to be added to the secondary to maintain an output of 300V

a)    Calculate the power in the load.Power = V x IP = 300 x 8.6=2580 WP= 2580W b)    If the primary voltage is 400V calculate the primary currentREMEMBER we do the opposite of the above equation (divide not multiply)2580/ 400 = IpIp= 6.5A

a)    The secondary is wound with 340 turns, how many turns are there on the primary. Primary voltage= Vp         Vp= 400         Vs= 300            VP                  Np                   Vs          =       Ns                          VP x Ns      =NP      Ns= Secondary turns= 340             Vs                     400          NP           400 x  340               = Np 453 (turns on primary)                     300          340           300  

  d) The primary voltage falls to 391 volts. How many turns need to be added to the secondary to maintain an output of 300V   Vp= 391 Vs=300 Np=453 Ns=?     Ns=     Vs x Np                     300 x 453                     Ns= 348 turns                      Vp                              391 453-348= 105 turns to be added

A 33kV to 11kV transformer carries 4.54A in the primary winding. There are 425 turns on the secondary winding. a)    Calculate the secondary current.  To calculate the secondary current-  Step one: identify what we already know.  Step two: we have Ns, Ip, Vp and Vs.  Step three: we know Vp=Is                                          Vs=Ip Step four: so we have to re arrange the formula to find Is Step 5: our new equation is Vp x IP                                                Vs Step 6: 33kV x 454A =   33 (x 10 to the power of 3) x 4.54                  11KV                  11(x 10 to the power of 3) Step 7: Is =13.62A

a)    Calculate the number of turns on the primary. Step one: identify what we already know- write out the formula! Step two: write out that – Ns= 425 turns Np=? Vp= 33kV Vs= 11kV Ip= 4.54 A Is= 13.62A    Step three: we know that Np = Vp = Is                                                     Ns =  Vs  =Ip           Step four: in order to find Np we have to rearrange the formula                            Vp xNs    = 33 x 425        = 1,275                                 Vs              11                                                      Step five: Np = 1,275

C)    The power consumed in the secondary. Step one: Formula for power is V x I                  So formula for secondary power is Vs x Is Step two: Vs x Is =Ps   11 x 13.62 (times voltage to 10 to the power of 3) Step three: Ps =14982W- 149.82kW  

D) The input voltage rises to 33.156kV, how many turns have to be removed from the secondary winding to maintain the output at 11KvVP= 33.156kVVs= 11kVNp= 1275 turnsNs=?Ip= 4.54Is=13.62AWork out what formula we need! Use previous questions for help if you need it. We know to find the number of primary turns we use                                                                                        Vp x Ns                                                                                    Vs           To find out number of secondary turns we can think about the above equation and invert   it.                                                                                                    Vs x Np       = 11(x 10 to the power of 3) x 1275                               Vp                           33.156(x 10 to the power of 3)   = 423 turns                      Ns =423 turn so 2 turns need to be taken away

Question1  Np  Vp     Is Ns = Vs =  Ip A power station transformer converts 25kv to 400kv ready for transmission via the national grid. a)    Calculate the turns ratio of the transformer. To calculate the turns ratio of a transformer you divide Vs by Vp 400kV = 1:16  25kV

b) Calculate the primary current if the secondary current is 25A.       To calculate the primary current we have to establish what we know:       Vp= 25kV       Vs= 400kV       Is= 25A       Ip=400A    We also know that Vp  is the same as Is   so 25kV                                Vs                          Ip       400kV Therefore Ip = 400kV

C)    The power consumed in the secondary. Power =V x I Secondary power = Vs x Is                                  400 x 25 = 100000kW To change from kW to MW you have to divide the answer by 1000 (or 10to the power -4) = 10MWd) The output power measured by a wattmeter is found to be 9.75MW what is the efficiency of the transformer.       Efficiency = output / Input       Efficiency= 9.75 / 10      Efficiency =0.975      Efficiency %= 97.5%