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Q.

A  750Hz, 20V source is connected to a resistance of  100Ω, an inductance of 0.1803 Henry and a capacitance of 10 microfarad, all in series. The time in which the resistance (thermal capacity =2joul/0C) will get heated by  100C is 

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a

7.8 min 

b

5.8 min

c

2.8 min

d

4.8 min 

answer is A.

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Detailed Solution

ω=2πf=2×3.14×750    XL=ωL=2×3.14×750×0.1803 =849.2Ω     XC=1ωC=12×3.14×750×10×106  =21.2Ω       XLXC=849.221.2828Ω           Z=R2+(XLXC)2  =(100)2+(828)2=834Ω

but   P=VrmsIrmscosϕ     =Vrms2Z2R=(20)2(100)(834)20.0575w

And total energy Lost= Heat produced in resistor 

Pt=mSΔθ      t=msΔθP =2×100.0575=348s5.8min

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