TS EAMCET 6-Aug-2021 Shift 1 Question Paper

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=‌

XC

Z

Vmax=‌

1000

1000√2

×20mV
‌‌=√2×10mV
‌‌=14.14mV