JEE Main · 2010 · 11 Apr 2010 · Offline · Alternating Current

In a series LCR circuit R =

200Ω

and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by

3 0^{\circ}

. On taking out the inductor from the circuit the current leads the voltage by

3 0^{\circ}

. The power dissipated in the LCR circuit is

(A)305 W
(B)210 W
(C)Zero W
(D)242 W✓ Correct

Answer: (D)

Solution

Removing the capacitor leaves

R

and

L

and the current lags by

3 0^{\circ}

; removing the inductor leaves

R

and

C

and the current leads by

3 0^{\circ}

. The two angles being equal tells us the inductive and capacitive reactances are equal,

X_{L} = X_{C}

, which is exactly the resonance condition for the full circuit. At resonance the reactances cancel, the impedance is just

R

and the power factor is one, so all the source power is dissipated in

R

P = \frac{V ^{2}}{R} = \frac{( 220 ) ^{2}}{200} = 242

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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JEE Main · 2010 · 11 Apr 2010 · Offline · Alternating Current

Solution

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