A pure inductor of 50.0 mH is connected to a | vedclass

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

Question

(C) Step $1$: Given values:$V = 220 \ V$$L = 50 \ mH = 50 	imes 10^{-3} \ H$$
u = 50 \ Hz$Step $2$: The inductive reactance $X_L$ is given by $X_L =

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) Step $1$: Given values:$V = 220 \ V$$L = 50 \ mH = 50 	imes 10^{-3} \ H$$
u = 50 \ Hz$Step $2$: The inductive reactance $X_L$ is given by $X_L = \omega L = 2 \pi 
u L$.Step $3$: The rms current $I$ is given by $I = \frac{V}{X_L} = \frac{V}{2 \pi 
u L}$.Step $4$: Substituting the values:$I = \frac{220}{2 	imes 3.14159 	imes 50 	imes 50 	imes 10^{-3}}$$I = \frac{220}{15.70795} \approx 14.005 \ A$.Thus,the rms current is approximately $14 \ A$.

$A$ pure inductor of $50.0 \ mH$ is connected to a source of $220 \ V$. The frequency of the source is $50 \ Hz$. The rms current in the circuit (in $A$) will be $......$.

Similar Questions

$A$ coil has an inductance of $2 \ H$. The ratio of its reactance when it is connected first to an $A.C.$ source and then to a $D.C.$ source is:

If the capacitance of a pure capacitor in a $DC$ circuit is $1 \ F$,then what will be its capacitive reactance?

An alternating voltage $V = 300\sqrt{2} \sin(100t)$ is connected to a $1 \ \mu F$ capacitor through an $AC$ ammeter. The reading of the ammeter will be.....$mA$.

$A$ choke coil is used to control:

Discuss power in an $AC$ circuit containing only a capacitor.

Vedclass Test Series

Exam Paper Generator

Online Exam Module