The $ r.m.s$. value of an ac of $ 50 Hz$ is $10 amp$. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
$2×10^{-2}$ $sec$ and $14.14$ $ amp$
$1 ×10^{-2}$ $sec$ and $7.07$ $ amp$
$5 × 10^{-3}$ $sec$ and $7.07 $ $amp$
$5 × 10^{-3}$ $sec$ and $14.14$ $ amp$
The voltage of an $ac$ source varies with time according to the equation $V = 100\sin \,\left( {100\pi t} \right)\cos \,\left( {100\pi t} \right)$ where $t$ is in seconds and $V$ is in volts. Then
If an alternating voltage is represented as $E = 141\,sin\, (628\,t),$ then the rms value of the voltage and the frequency are respectively
An alternating voltage is represented as $E = 20\,sin \,300t.$ The average value of voltage over one cycle will be.......$V$
A $40$ $\Omega$ electric heater is connected to a$ 200 V, 50 Hz$ mains supply. The peak value of electric current flowing in the circuit is approximately......$A$
For the $RC$ circuit shown, the resistance is $R = 10.0\ W$, the capacitance is $C = 5.0\ F$ and the battery has voltage $\xi= 12$ volts . The capacitor is initially uncharged when the switch $S$ is closed at time $t = 0$. At some time later, the current in the circuit is $0.50\ A$. What is the magnitude of the charge across the capacitor at that moment?.......$µC$