$A$ power transformer is used to step up an alternating $e.m.f.$ of $220\,V$ to $11\,kV$ to transmit $4.4\,kW$ of power. If the primary coil has $1000$ turns,what is the current rating of the secondary (in $,A$)? Assume $100\%$ efficiency for the transformer.

The ratio of secondary to primary turns in a transformer is $3 : 2$. If the power output is $P$,then the input power,neglecting all losses,must be equal to:

The primary of a transformer has $100$ turns and operates at $100 \,V - 200 \,W$. The number of turns in the secondary,if the output voltage is $2000 \,V$,is:

Primary side of a transformer is connected to $230 \ V, 50 \ Hz$ supply. The turns ratio of primary to secondary winding is $10:1$. The load resistance connected to the secondary side is $46 \ \Omega$. The power consumed in it is: (in $W$)

$A$ transformer has $120$ turns in the primary coil and carries $5 \text{ A}$ current. Input power is $1 \text{ kW}$. To have $560 \text{ V}$ output, the number of turns in the secondary coil will be:

$A$ power transmission line feeds input power at $2.3 \text{ kV}$ to a step-down transformer with its primary winding having $3000$ turns. The output power is delivered at $230 \text{ V}$ by the transformer. The current in the primary of the transformer is $5 \text{ A}$ and its efficiency is $90 \%$. The winding of the transformer is made of copper. The output current of the transformer is . . . . . . $A$.