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 $A$)?

An ideal transformer with a purely resistive load operates at $12\,kV$ on the primary side. It supplies electrical energy to a number of nearby houses at $120\,V$. The average rate of energy consumption in the houses served by the transformer is $60\,kW$. The value of the resistive load $(R_s)$ required in the secondary circuit will be $.........\,m\Omega$.

$A$ 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:

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:

An ideal transformer converts $220 \ V$ $AC$ to $3.3 \ kV$ $AC$ to transmit a power of $4.4 \ kW$. If the primary coil has $600$ turns,then the alternating current in the secondary coil is:

An $AC$ source of internal resistance $10^3 \Omega$ is connected to a transformer. The ratio of the number of turns in the primary to the number of turns in the secondary to match the source to a load resistance of $10 \Omega$ is