Question

An AC source of emf E =200sin(100πt) is connected across an inductor having resistance
1002 and self inductance 2H. Calculate-
(i) Frequency of AC.
(ii) Total impedance of the circuit.
(iii) Peak value of current flowing through the
circuit. ​

Answer 1

Answer:

ANSWER

a) ε is ϕ ahead of i

because i and R is in phase and i.e., −Z and ε are in phase as ε=ε

0

sinwt

from diagram

i=

−Z

ε

0

sin(ωt−ϕ)

where, ϕ=tan

−1

R

X

L

b) i=1.0A (given)

V

R

=160V=iR⇒R=

1

160

=160

V

L

=120V=iX

L

⇒X

L

=

1

120

=120

V

net

=

V

R

2

+V

L

2

=

(160)

2

+(120)

2

=

25600+14400

=200 : effective value of apphed voltage

Z=

R

2

+X

L

2

=

(160)

2

+(120)

2

=200 : Impedence

C) When direct current is passed through the circuit then R will be zero)

at t=ω, inductor will act as plane wire and potential drop will only act occur at R

(i.e.,i=

R

ε

)

For intermediate time

E=iR+

dt

Ldi

where i will vary accordingly

solution