# transformer current in primary coil

• Feb 25th 2008, 07:57 PM
rcmango
transformer current in primary coil
A step-down transformer (turns ratio = 1:8) is used with an electric train to reduce the voltage from the wall receptacle to a value needed to operate the train. When the train is running, the current in the secondary coil is 1.9 A.

What is the current in the primary coil?
in Amperes

if the formula for current in the secondary coil is: A * (ratio of primary coil/secondary coil) = current

then why can't we just use A* 1/8 = 1.9

multiply by 8 and get the amps for the first coil which is 15.2 A
• Feb 26th 2008, 08:42 AM
earboth
Quote:

Originally Posted by rcmango
A step-down transformer (turns ratio = 1:8) is used with an electric train to reduce the voltage from the wall receptacle to a value needed to operate the train. When the train is running, the current in the secondary coil is 1.9 A.

What is the current in the primary coil?
in Amperes
if the formula for current in the secondary coil is: A * (ratio of primary coil/secondary coil) = current

then why can't we just use A* 1/8 = 1.9
multiply by 8 and get the amps for the first coil which is 15.2 A

I hope this reply doesn't come too late.

$n_p, n_s$ are the turns in the primary or secondary coils.
$U_p, U_s$ are the voltages in the primary or secondary coils.
$I_p, I_s$ are the currents in the primary or secondary coils.
$E_p, E_s$ are the electrical energies in the primary or secondary coils.

Then you know:

$\frac{U_p}{U_s}= \frac{n_p}{n_s}$

You know by the law of conservation of energy that

$E_p = E_s$ . That means:

$U_p \cdot I_p = U_s \cdot I_s~\iff~\frac{U_p}{U_s} = \frac{I_s}{I_p}~\iff~ \frac{n_p}{n_s} = \frac{I_s}{I_p}$ .

$\frac{n_p}{n_s} = \frac{8}{1}$ . And therefore you get:
$\frac{8}{1} = \frac{I_s}{I_p}~\implies~I_s = \frac{8}{1} \cdot I_p$ . Plug i the value you know and solve for $I_p$.
I've got $I_p = 0.2375\ A$