Transformer and power question mutiple choice

[Clara Chung]

Homework Statement

The turns ratio of a transformer is 20:1 step down and the primary coil is connected to a 220V a.c. supply. If the secondary coil is connected to a '36W, 15V' light bulb and the efficiency of the transformer is 80%,find the current in the primary coil.

Homework Equations

P=IV, Ns/Np=Ip/Is

The Attempt at a Solution

I calculate by 36/0.8 = 45 W, 45/220 = 0.204A, which is the answer of D, but the correct answer is B, which is 0.11A, Am I wrong or the is the question wrong?[/B]

 

[cnh1995]

Homework Statement

The turns ratio of a transformer is 20:1 step down and the primary coil is connected to a 220V a.c. supply. If the secondary coil is connected to a '36W, 15V' light bulb and the efficiency of the transformer is 80%,find the current in the primary coil.

Homework Equations

P=IV, Ns/Np=Ip/Is

The Attempt at a Solution

I calculate by 36/0.8 = 45 W, 45/220 = 0.204A, which is the answer of D, but the correct answer is B, which is 0.11A, Am I wrong or the is the question wrong?[/B]

36W is the power rating of the bulb at 15V. The bulb is not getting 15V in this situation, so the power is not 36W. What is the actual operating voltage and resistance of the bulb?

 

[Clara Chung]

36W is the power rating of the bulb at 15V. The bulb is not getting 15V in this situation, so the power is not 36W. What is the actual operating voltage and resistance of the bulb?

That's all the information given. I think the resistance of the bulb can be found by 15^2 / 36= 6.25 ohm

 

[cnh1995]

That's all the information given. I think the resistance of the bulb can be found by 15^2 / 36= 6.25 ohm

Right. Now what's the voltage across the bulb?

 

[Clara Chung]

Right. Now what's the voltage across the bulb?

Voltage 220/20 = 11V, but isn't there energy lost?

 

[cnh1995]

but isn't there energy lost?

Yes. That will reduce the secondary voltage. But I think this is how it will be resolved..
If the secondary is kept open and you applied 220V across the primary, you'll get 11V across the secondary, which is the maximum OC voltage, or secondary induced emf E2. Now, if the secondary is loaded, the load voltage will drop. In practice, load voltage should be equal to the secondary induced voltage. A tap changer is used in the primary which will reduce the number of primary turns and hence, secondary induced emf will be increased such that load voltage remains 11V.

 

[Clara Chung]

Yes. That will reduce the secondary voltage. But I think this is how it will be resolved..
If the secondary is kept open and you applied 220V across the primary, you'll get 11V across the secondary, which is the maximum OC voltage, or secondary induced emf E2. Now, if the secondary is loaded, the load voltage will drop. In practice, load voltage should be equal to the secondary induced voltage. A tap changer is used in the primary which will reduce the number of primary turns and hence, secondary induced emf will be increased such that load voltage remains 11V.

thanks a lot with those extra information!DDD now I get the answer by 11^2 / 6.25 =19.4 W, 19.4/0.8 =24.2 W ,24.2/220 = 0.11A :D

 

[cnh1995]

thanks a lot with those extra information!DDD now I get the answer by 11^2 / 6.25 =19.4 W, 19.4/0.8 =24.2 W ,24.2/220 = 0.11A :D

I'm not sure if my above reasoning is true for such a small system but in large power systems, this is how the load voltage is stabilized. Maybe this transformer has no copper loss and only iron loss is present. So this will not cause any voltage drops and you'll get 11V across the load.