Request a call back

A resistance of 600 , an inductor of 0.4 H and a capacitor of 0.01 uF are connected in series to an AC source of variable frequency. Find the frequency of AC source for which current in the circuit is maximum. Also calculate the band width and quality factor for the circuit.

Asked by shivakumarshreyas24 | 01 Mar, 2020, 08:12: AM
frequency of AC source when current is maximum, ωo = 2πf = 1/ (LC)1/2

Hence

Bandwidth  2Δω = R/L = 600/.4 = 1500 Hz = 1.5kHz

Quality Q = ( ωo L ) / R = ( 2516 × 0.4 ) / 600 = 1.68
Answered by Thiyagarajan K | 01 Mar, 2020, 12:35: PM

## Concept Videos

CBSE 12-science - Physics
Asked by shivakumarshreyas24 | 01 Mar, 2020, 08:12: AM
CBSE 12-science - Physics
Asked by khushimassey437 | 31 May, 2019, 08:41: AM
CBSE 12-science - Physics
Asked by manasvijha | 19 Mar, 2019, 07:17: PM
CBSE 12-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
CBSE 12-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
CBSE 12-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM
CBSE 12-science - Physics
Asked by Topperlearning User | 04 Jun, 2014, 01:23: PM