LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
in Physics by (30.0k points)
closed by
The coil of the inductor element of 0.5 H is connected to an alternating source of frequency 50 Hz and its impedance will be -
1. 25 Ω
2. 100 Ω
3. 157 Ω
4. 622 Ω

1 Answer

0 votes
by (54.3k points)
selected by
Best answer
Correct Answer - Option 3 : 157 Ω


  • Reactance: It is basically the inertia against the motion of the electrons in an electrical circuit.
  • The unit of reactance is the ohm. In an electrical circuit.

\(⇒ Reactance= \frac{V}{I}\)

Where V = potential difference and I = current

  • Impedance is a combination of resistance and reactance. It is essentially everything that resists the flow of electrons within an electrical circuit.
  • Impedance of inductive circuit is given as

Znet =  jωL


'Z = 2 π f L

Where f is frequency of AC Circuit, L is inductance



frequency f = 50 Hz

Inductance :L = 0.5 H

'Z = 2 π f L = 2 × π × 0.5 × 50 = 157. 07

So, Z ≈ 157 Ω 

So, the correct option is 157 Ω.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.