The circuit shown below will:
The circuit shown below will: Correct Answer Pass higher frequencies
Concept:
The inductive reactance is given by:
XL = jωL
ω = frequency applied
L = Inductance
Analysis:
For ω (Input frequency) = 0 rad/s:
The inductive reactance will be:
XL = j(0)L = 0 Ω (Short circuit)
The circuit is drawn as:
[ alt="F2 Shubham Bhatt 3.3.21 Pallavi D13" src="//storage.googleapis.com/tb-img/production/21/03/F2_Shubham%20Bhatt_3.3.21_Pallavi_D13.png" style="width: 207px; height: 128px;">
The output voltage is therefore 0 V.
For ω (Input frequency) = ∞ rad/s:
The inductive reactance will be:
XL = j(∞)L = ∞ Ω (Open circuit)
The circuit is drawn as:
[ alt="F2 Shubham Bhatt 3.3.21 Pallavi D14" src="//storage.googleapis.com/tb-img/production/21/03/F2_Shubham%20Bhatt_3.3.21_Pallavi_D14.png" style="width: 212px; height: 129px;">
The output voltage is therefore 10 V.
Conclusion: The circuit does not allow low frequencies to pass and only passes high frequencies. Option (1) is correct.
