Given that, input AC voltage, Vi(t)=20sin(105t)mV As, V=Vmaxsinωt
Comparing Eqs. (i) and (ii), we get ω=105rad∕s,Vmax=20mV=20×10−3V The Vi is distributed across R and C. by VR and VC, as phasor diagram. Reactance of capacitor, XC=
1
ωC
=
1
105×10−8
Ω
=103Ω=1000Ω ∴ Now, impedance of R−C series circuit. Z=√R2+XC2=√(1000)2+(1000)2=1000√2Ω Now, voltage across capacitor, VC=V0=
