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):

The number of turns in the primary and secondary coils of a transformer are $50$ and $200$ respectively. If the current in the primary coil is $4 \,A$, then the current in the secondary coil is: (in $\,A$)

"In a step-down transformer, the output current decreases." This statement is true or false?

An $80 \text{ W}$ lamp is connected to the secondary of a step-down transformer, where the primary is connected to an $AC$ mains of $220 \text{ V}$. Assuming the transformer to be ideal, the current in the primary winding is nearly: (in $\text{ A}$)

$A$ transformer has $100$ turns in the primary coil and carries $8$ $A$ current. If input power is $1$ $kW$, the number of turns in the secondary coil to have $500$ $V$ output will be

$A$ step-down transformer has a $N_{S}$ to $N_{P}$ ratio of $1: 20$. If $8 \, V$ are developed across a $0.4 \, \Omega$ secondary load, the primary current will be: (in $ \, A$)