A 750 Hz, 20 V (rms) source is connected to a resistance of `100 Omega`, an inductance of 0.1803 H and a capacitance of `10 mu F` all in series. The time in which the resistance (heat capacity `2 J//.^@C`) will get heated by `10^@C`. (assume no loss of heat to hte surroundings) is close to :

Explanation

The imepedance of the circuit
`Z=sqrt(R^2+(X_L-X_C)^2)=sqrt(R^(2)+{(2pifL)-1/((2pi fC))}^(2))`
`sqrt((100)^(2)+{(2xx3.14 xx 750 xx 0.1803)-1/((2 xx 3.14 xx 750 xx 10^-5))}^(2))`
`=834Omega`
In case of an `AC`,
`ltlt P gtgt =V_("rms") i_("rms") cosphi=V_("rms")(V_("rms")/Z)(R/Z)=(V_("rms")/Z)^2R`
`=(20/834)^2xx100=0.0575J//s`
Now, `ltlt P gtgt xxt=S Delta theta`
Here S=thermal capacity
`:. t=(S Delta theta)/(langle P rangle)=(2xx10)/0.0575=348s`.