$A$ 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:

$A$ transformer is used to

$A$ step-up transformer operates on a $230\, V$ line and supplies a load of $2\, A$. The ratio of the primary and secondary windings is $1 : 25$. The current in the primary is.... (in $, A$)

$A$ 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 . . . . . . $A$.

Using the equation of power for an ideal transformer,prove $\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$.

$A$ transformer is used to step up an alternating e.m.f. of $220 \ V$ to $4.4 \ kV$ to transmit $6.6 \ kW$ of power. The primary coil has $1000$ turns. The current rating of the secondary coil is (Transformer is ideal): (in $A$)