For a series LCR circuit with $\mathrm{R}=100 \Omega, \mathrm{~L}=0.5 \mathrm{mH}$ and $\mathrm{C}=0.1 \mathrm{pF}$ connected across $220 \mathrm{~V}-50 \mathrm{~Hz}$ AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :

(C) $\approx 90^{\circ}$, predominantly capacitive circuit

(A) $\approx 90^{\circ}$, predominantly inductive circuit

(B) $0^{\circ}$, resonance circuit

(D) $0^{\circ}$, resistive circuit

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Alternating Current

For a series LCR circuit with

R = 100Ω, L = 0.5 mH

and

C = 0.1 pF

connected across

220 V - 50 Hz

AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :

Answer: (C)

With

ω = 2 π f = 100 π rad s^{- 1}

the two reactances are

X_{L} = ω L = 100 π \times 0.5 \times 1 0^{- 3} \approx 0.16 Ω

X_{C} = \frac{1}{ω C} = \frac{1}{100 π \times 0.1 \times 1 0 ^{- 12}} \approx 3.2 \times 1 0^{10} Ω

X_{C}

exceeds both

R

and

X_{L}

by many orders of magnitude, so the impedance is almost entirely capacitive and

tan ϕ = \frac{X _{C} - X _{L}}{R} \approx 3 \times 1 0^{8}

The phase angle is therefore about

9 0^{\circ}

with the current leading the supply voltage, so the circuit is predominantly capacitive.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option