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A series LCR circuit is connected to a 200 {~V}, 300 rads ^{-1

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Solving time: 3 mins

A series LCR circuit is connected to a $200 V, 300$ rads $^{- 1}$ source. The resistance in the circuit is $100Ω$ . When the capacitor is removed from the circuit, the current in the circuit lags the source voltage by $6 0^{\circ}$ . When only the inductance is removed, the current leads the voltage by $6 0^{\circ}$ . Calculate the current and power dissipated in the original $L CR$ circuit.

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Book:

Physics for JEE Main and Advanced Magnetism and EMI (McGraw Hill)

Text solutionVerified

When

C

is removed, the circuit becomes an

L R

circuit with

tan ϕ = \frac{X _{L}}{R} \Rightarrow X_{L} = R tan ϕ = 1003 Ω

when

L

is removed, the circuit becomes a

CR

circuit with

tan ϕ = \frac{X _{C}}{R} \Rightarrow X_{C} = R tan ϕ = 1003 Ω

when

L, C

and

R

all are present the circuit is in resonance as

X_{L} = X_{R}

∴ Z = R = 100Ω

∴ I = \frac{V}{Z} = \frac{200}{100} = 2 A

P = V I cos ϕ = 200 \times 2 \times 1 = 400 W

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Camila Discussed

A series LCR circuit is connected to a

200 V, 300

rads

^{- 1}

source. The resistance in the circuit is

100Ω

. When the capacitor is removed from the circuit, the current in the circuit lags the source voltage by

6 0^{\circ}

. When only the inductance is removed, the current leads the voltage by

6 0^{\circ}

. Calculate the current and power dissipated in the original

L CR

circuit.

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Question Text	A series LCR circuit is connected to a $200 V, 300$ rads $^{- 1}$ source. The resistance in the circuit is $100Ω$ . When the capacitor is removed from the circuit, the current in the circuit lags the source voltage by $6 0^{\circ}$ . When only the inductance is removed, the current leads the voltage by $6 0^{\circ}$ . Calculate the current and power dissipated in the original $L CR$ circuit.
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	91

Text solutionVerified

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