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A 50 Hz AC signal is applied in a circui...

A 50 Hz AC signal is applied in a circuit of inductance of `(1//pi)`H and resistance `2100Omega.` The impedance offered by the circuit is

A

`1500 Omega`

B

`1700 Omega`

C

`2102 Omega`

D

`2500 Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
Impedance, `Z= sqrt(R^(2)+X_(L)^(2)), X_(L)=omegaL=2pifL`
Given, `R=2100Omega`, f=50Hz, `L=1/piH`
`rArr Z= sqrt((2100)^(2) + (2 xx 50)^(2)) = sqrt((2100)^(2) + (100)^(2))`
`rArr Z=2102 Omega`
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