In an ideal transformer,the turns ratio is $\frac{N_p}{N_S} = \frac{1}{2}$. The ratio $V_s : V_p$ is equal to (the symbols carry their usual meaning):

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 has an efficiency of $80 \%$ and works at $10 \ V$ and $4 \ kW$. If the secondary voltage is $240 \ V$, then the current in the secondary coil is: (in $A$)

$A$ current of $5 \,A$ is flowing at $220 \,V$ in the primary coil of a transformer. If the voltage produced in the secondary coil is $2200 \,V$ and $50 \%$ of power is lost, then the current in the secondary will be (in $A$)

An ideal transformer has a turns ratio of $10$. When the primary is connected to $220 \ V$,$50 \ Hz$ as a source,the power output is

$A$ small town with a demand of $900 \text{ kW}$ of electric power at $220 \text{ V}$ is situated $20 \text{ km}$ away from an electric power generating station. The two-wire line has a resistance per unit length of $5 \times 10^{-4} \text{ } \Omega \text{ m}^{-1}$. The town gets power from the line through a $45000 \text{ V}$ to $220 \text{ V}$ step-down transformer at a substation in the town. The line power loss in the form of heat is: (in $text{ kW}$)