In the circuit shown in the figure, the ac source gives a voltage V = 24cos(2000t). Neglecting source resistance, the voltmeter and ammeter reading will be
Z=√(R)2+(XL−Xc)2
R=12Ω,XL=wL=2000× 5× 10−3=10Ω
Xc=1wC=12000× 50× 10−6=10Ω i.e.Z = 10Ω
Maximum current i0=V0Z=2412=2A
Hence irms=2√2=1.4A
and Vrms=6× 1.41=8.46 V