An alternating voltage $v\left( t \right) = 220\,\sin \,100\pi l\,volt$ is applied to a purely resistive load of $50\,\Omega $ . The time taken for the current to rise from half of the peak value of the peak value is.....$ms$
An alternating voltage $v\left( t \right) = 220\,\sin \,100\pi l\,volt$ is applied to a purely resistive load of $50\,\Omega $ . The time taken for the current to rise from half of the peak value of the peak value is.....$ms$
- [JEE MAIN 2019]
- A
$2.2$
- B
$3.3$
- C
$5$
- D
$7.2$
Similar Questions
The mean and $rms$ value of an alternating voltage for half cycle as shown in figure are respectively:-
The mean and $rms$ value of an alternating voltage for half cycle as shown in figure are respectively:-
A $280$ ohm electric bulb is connected to $200V$ electric line. The peak value of current in the bulb will be
A $280$ ohm electric bulb is connected to $200V$ electric line. The peak value of current in the bulb will be
In $ac$ circuit when $ac$ ammeter is connected it reads $i$ current if a student uses $dc$ ammeter in place of $ac$ ammeter the reading in the $dc$ ammeter will be:
In $ac$ circuit when $ac$ ammeter is connected it reads $i$ current if a student uses $dc$ ammeter in place of $ac$ ammeter the reading in the $dc$ ammeter will be:
An AC source is rated $222 \,V , 60 \,Hz$. The average voltage is calculated in a time interval of $16.67 \,ms$. It
An AC source is rated $222 \,V , 60 \,Hz$. The average voltage is calculated in a time interval of $16.67 \,ms$. It
the reason why do we preferred an $a.c.$ voltage instead of $d.c.$ voltage.
the reason why do we preferred an $a.c.$ voltage instead of $d.c.$ voltage.