Average power associated with an ideal induct | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $1$. Ideal Inductor:- The voltage and current in an ideal inductor are out of phase by $90^{\circ}$,with the current lagging the voltage by $90^{\

Vedclass · Accepted Answer

zero,zero

Vedclass · Answer

(C) $1$. Ideal Inductor:- The voltage and current in an ideal inductor are out of phase by $90^{\circ}$,with the current lagging the voltage by $90^{\circ}$.- The instantaneous power is given by $p(t) = v(t) 	imes i(t)$.- Over a complete cycle,the positive and negative power values cancel out,resulting in an average power of zero.$2$. Ideal Capacitor:- In an ideal capacitor,the voltage and current are also out of phase by $90^{\circ}$,but the current leads the voltage by $90^{\circ}$.- Similar to the inductor case,the instantaneous power alternates between positive and negative,resulting in an average power of zero over a complete cycle.

Average power associated with an ideal inductor and ideal capacitor over a complete cycle of $a.c.$ is respectively

Similar Questions

The capacitive reactance of a capacitor $C$ is $X \ \Omega$. Both the frequency of the $A.C.$ supply and the capacitance of the capacitor are doubled. The new capacitive reactance will be:

An $AC$ supply gives $30 \ V$ $rms$ which passes through a $10 \ \Omega$ resistance. The power dissipated in it is ...... $W$.

$A$ $15.0 \mu\text{F}$ capacitor is connected to a $220 \text{ V}$, $50 \text{ Hz}$ source. The capacitive reactance of the circuit is . . . . . . $\Omega$.

$A$ capacitor of $50 \mu F$ is connected to an a.c. source $e = 220 \sin(50t)$ (where $e$ is in volts and $t$ is in seconds). The value of the peak current is:

Explain $AC$ voltage applied to a resistor and explain it with a necessary graph.

Vedclass Test Series

Exam Paper Generator

Online Exam Module