# 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 -

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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

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Correct Answer - Option 3 : 157 Ω

Concept:

• 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.