1.

A series resonant LCR circuit has a quality factor (Q-factor)=0.4. If `R=2k Omega, C=0.1 mu F` then the value of inductance isA. `0.1 H`B. `0.064 H`C. 2 HD. 5 H

Answer» Correct Answer - B
Quality factor `Q = (1)/(R )sqrt((L)/(C ))` or `(L)/(C )=(QR)^(2)`
Here, `Q = 0.4, R = 2 k Omega= 2xx10^(3)Omega`
`C=0.1 mu F=0.1xx10^(-6)F`
`therefore L=(QR)^(2)C`
`therefore L =(0.4xx2xx10^(-3))^(2)xx0.1xx10^(-6)=0.064 H`


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