Question
Download Solution PDFThe coil of the inductor element of 0.5 H is connected to an alternating source of frequency 50 Hz and its impedance will be -
Answer (Detailed Solution Below)
Option 3 : 157 Ω
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DSSSB TGT Hindi Female 4th Sep 2021 Shift 2
16.3 K Users
200 Questions
200 Marks
120 Mins
Detailed Solution
Download Solution PDFConcept:
- 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
or
'Z = 2 π f L
Where f is frequency of AC Circuit, L is inductance
Calculation:
Given
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.
Last updated on May 12, 2025
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