How much current is drawn by the primary coil of a transformer,which steps down $220 V$ to $55 V$ to operate a device with an impedance of $275 \Omega$ (in $A$)?

$A$ generator produces a current of $100 \, A$ at $4000 \, V$. The voltage is stepped up to $2 \times 10^5 \, V$ by a transformer before being sent on a high voltage transmission line of resistance $50 \, \Omega$. The percentage of power loss in the transmission line is (in $\%$)

$A$ power transmission line feeds input power at $2300\,V$ to a step-down transformer with its primary windings having $4000$ turns,giving the output power at $230\,V.$ If the current in the primary of the transformer is $5\,A,$ and its efficiency is $90\%,$ the output current would be......$A$

The number of turns in the primary and the secondary coils of a transformer are $1000$ and $3000$ respectively. If $80 \,V$ a.c. is applied to the primary coil of the transformer,then the potential difference per turn of the secondary coil would be: (in $\,V$)

$A$ 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:

Find the current through the primary coil $(P)$ of the transformer shown below. (in $A$)