Thread: Electrical current problem involving a step-down transformer.

1. Electrical current problem involving a step-down transformer.

I have a little problem with this multi-part question, I think I have some of the answers but other parts of the questions throw me a little bit

P1):
Power lines carry electricity at about 120,000 volts. If a power line were stepped down by a transformer to about 120, volts how many turns would there be on the secondary coil if there were 1000 turns on the primary?

A1):
There should only be one turn on the secondary since: Primary voltage/turns in primary coil=secondary voltage/turns in secondary coil. If primary voltage= 120,000, turns in primary coil=1000 and secondary voltage=120 then 120,000/1000=120/x and x=1.

P2):
If the secondary coil in part 1 is connected to a resistor with a resistance of $\displaystyle 10\Omega$, what is the current of the second coil?

A2):
By Ohm's law current equals:voltage/resistance. If voltage=120 and resistance= 10 then 120/10=12. Therefore, there are 12A of current flowing through the secondary coil.

P3):
What is the power in the secondary coil? show your calculations

A3):
Power=current*voltage. If current=12A and voltage=120 then: power=12*120=1440w=1.44Kw.

P4):
What is the current drawn by the primary coil?Show your calculations.

A4):
The question does not specify any resistor connected to the coil, therefore I will assume that the resistance on the coil is negligible so, if resistance=0 then current=0.
there is no current flowing across the primary coil.
(I'm not sure if I'm right on this one especially.)

Thank you!

2. Dear mathblaster,

You have been posting problems in science areas but you say you are in a crash program algebra 2 an d trig. I am interested in where you get these and I have answered a couple. Are you a high school student?

bjh

3. Yes, I am a high school student, I'm doing some self teaching in order to get my grades up.
I forgot to alter my sig(kinda silly of me really), I've already finished teaching myself algebra 2 and trig to a level that I feel I can pass with so I'm moving on to science--which I'm close to done with. I'm just trying to make sure I have my facts down and that I understand the subjects properly.

Cheers!

4. hello again mathblaster,

Where are you getting the questions? Do you have study materials and any guideance from a science teacher? I don't think that doing problems without study is the right course. Other responders may have inputs. Relative to your electrical question I am very hazy on this but I know that high voltage transmission lines are stepped down before the step down to single phase 220 v. I think that this last step is from 13000 v where I live.The 220 is two wire with a ground and each leg is 110v to ground. current returns to ground from primary and secondary magnetic cores. voltage between legs 220

bjh

5. I would agree with all your answers up to part 4. You were right to be a little hesitant. There must be power transferred from the primary coil to the secondary coil. There must also be current, or the transformer simply wouldn't operate. Since the secondary coil is dissipating 1.44kW, the primary coil must be delivering at least that much power to the secondary coil, or you'd have invented the perpetual motion machine. P = I V is true for any circuit component. If you assume negligible losses in the primary coil, then what current do you get?

6. You can invoke several transformer equations to solve for the current. Since you know current 2, you can simply say:
$\displaystyle N_p/N_s = I_s/I_p$
where p = primary and s = secondary.

You could also use the impedance transformation and do
$\displaystyle I_p = \frac{v_p}{z_{in}}$
where z_in is the effective impedance seen by anything hooked up to the primary coil of the transformer. You can calculate it like so:
$\displaystyle z_{in} = a^2Z_s$
where a = n_p/n_s and Z_s is the impedance directly hooked up to the secondary side's coil.

A third approach would be to use the conservation of energy that an ideal transformer is famous for:
$\displaystyle P_s = P_p$
or
$\displaystyle v_si_s=v_pi_p$
which as you can see uses the first rule i gave you... because you can put the voltages on one side and the currents on another side and recognize that v_p/v_s = n_p/n_s.
but you can use this idea a little more economically (save some computation) since you already calculated p_s:
$\displaystyle v_pi_p=P_s$
$\displaystyle i_p = \frac{P_s}{v_p}$