A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of `32 Omega`. Coil Q has a self inductance of 4.9 mH and a resistance of `68 Omega` in series. The frequency is adjusted so that maximum current flows in P and Q.

The impedance of Q at this frequency is

Options

A. `9.76 V, 8.92 V`
B. `6.29 V, 7. 96 V`
C. `7.70 V, 10.92 V`
D. `7.70 V,9. 76 V`

Answer

D

Explanation

As this circuit is series LCR circuit current will be maximum at resonance,

`therefore` Resonance frequency
`omega =(1)/(sqrt(LC))=(1)/(sqrt((4.9xx10^(-3))(10^(-6))))=(10^(5))/(7)"rad "s^(-1)`
and current `I_("max")=(epsilon_(0))/(R )=(10)/((32+68))=(1)/(10)A`
So the impedance for box P
`Z_(P)=[R_(1)^(2)+((1)/(omega C))^(2)]^(1//2)=[(32)^(2)+((7)/(10^(5)xx10^(-6)))^(2)]^(1//2)`
`=sqrt(5924)=77 Omega`
and for coil `Q, Z_(Q)=[R_(2)^(2)+(omega L)^(2)]^(1//2)`
`=[(68)^(2)+((10^(5))/(7)xx4.9xx10^(-3))^(2)]^(1//2)=sqrt(9524)=97.6 Omega`
and hence , `V_(P)=I_("max)Z_(P)=(1)/(10)xx(77)=7.70 V`
(potential drop across P)
and `V_(Q)=I_("max")Z_(Q)=(1)/(10)xx(97.6)=9.76 V`
(potential drop across Q)