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It is given in the fig that the AC changes as 2sin(100t)2sin(100t). So the angular frequency of the AC is omega=100"rad/s"ω=100rad/s
Now Capacitive reactance of the circuit X_C=1/(omega C)XC=1ωC,, where C is the unknown capacitance in the given RLC series circuit.
The inductive reactance of the circuit X_L=omega LXL=ωL,, where L=0.1HL=0.1H is the inductance in the given RLC series circuit.
Again it is also given that resistance connected in RLC series circuit is R=10Omega and the power factor is cos phi=1/sqrt2, where phi is the angle subtended by the phasor with R in the impedance triangle.
As cos phi=1/sqrt2 then phi = cos ^-1(1/sqrt2)=45^@
tanphi= abs(X_L-X_C)/R
=>tan45^@= abs(omegaL-1/(omegaC))/R
=>1= abs(100xx0.1-1/(100C))/10
=>10= abs(10-1/(100C))
If 10>1/(100C) then value of C will be impossible one. So inthis case 10<1/(100C) and the overall circuit reactance is capacitive giving a leading phase angle 45^@
So the equation becomes
=>10= -10+1/(100C)
=>1/(100C)=20
=>C=1/2000F=500/10^6F=500xx10^-6F=500muF which is option (3)
