The output of a step-down transformer is measured to be $24 \ V$,when connected to a $12 \ W$ light bulb. The value of peak current is $...... \ A$. (in $A$)

An $AC$ current is given by $I = I_{1} \sin \omega t + I_{2} \cos \omega t$. $A$ hot wire ammeter will give a reading of:

$A$ resistance of $20 \Omega$ is connected to a source of an alternating potential $V = 200 \sin (10 \pi t)$. If $t$ is the time taken by the current to change from the peak value to rms value,then $t$ is (in seconds).

Find the effective value of current $i = 2\sin(100\pi t) + 2\cos(100\pi t + 30^{\circ})$.

In Karnataka,the normal domestic power supply $AC$ is $220 \text{ V}, 50 \text{ Hz}$. Here $220 \text{ V}$ and $50 \text{ Hz}$ refer to:

The output voltage of a step-down transformer is measured to be $24 \text{ V}$,when connected to a $12 \text{ W}$ light bulb. The value of the peak current is . . . . . . .