A series AC circuit containing an inductor ( 20 mH ), a capacitor ( $120 \mu \mathrm{~F}$ ) and a resistor ( $60 \Omega$ ) is driven by an AC source of $24 \mathrm{~V} / 50 \mathrm{~Hz}$. The energy dissipated in the circuit in 60 s is:

(C) $5.17 \times 10^{2} \mathrm{~J}$

(A) $5.65 \times 10^{2} \mathrm{~J}$

(B) $2.26 \times 10^{3} \mathrm{~J}$

(D) $3.39 \times 10^{3} \mathrm{~J}$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Alternating Current

A series AC circuit containing an inductor ( 20 mH ), a capacitor (

120 μ F

) and a resistor (

60Ω

) is driven by an AC source of

24 V /50 Hz

. The energy dissipated in the circuit in 60 s is:

Answer: (C)

In an AC series

L C R

circuit the only element that actually dissipates energy is the resistor, so the energy lost in a time

t

E = P t = \frac{V ^{2}}{Z ^{2}} R t

, using the rms voltage. First pin down the two reactances:

X_{L} = ω L = 2 π (50) (0.02) = 2 π Ω \approx 6.3 Ω

, while

X_{C} = \frac{1}{ω C} = \frac{1}{2 π ( 50 ) ( 120 \times 1 0 ^{- 6} )} = \frac{1 0 ^{4}}{12 π} Ω \approx 265 Ω

. The impedance then follows from

Z^{2} = R^{2} + (X_{L} - X_{C})^{2} = 6 0^{2} + (2 π - \frac{1 0 ^{4}}{12 π})^{2}

. Feeding this into the energy expression with

V = 24 V

R = 60 Ω

and

t = 60 s

gives

E = \frac{( 24 ) ^{2}}{Z ^{2}} (60) (60) \approx 5.17 \times 1 0^{2} J

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option