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In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit is
- a)242 W
- b)305 W
- c)210 W
- d)0 W
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In a series LCR circuit, R = 200Ω and the voltage and frequency of t...
Given:
R = 200Ω
Voltage (V) = 220 V
Frequency (f) = 50 Hz
Current lags behind the voltage by 30°:
When the capacitance is present in the circuit, the current lags behind the voltage by 30°. This implies that the circuit is capacitive in nature.
Current leads the voltage by 30°:
When the inductor is present in the circuit, the current leads the voltage by 30°. This implies that the circuit is inductive in nature.
Power dissipated in the LCR circuit:
The power dissipated in the circuit can be calculated using the formula: P = IV cos(θ), where P is the power, I is the current, V is the voltage, and θ is the phase angle between the current and voltage.
Power dissipated in the circuit with capacitance:
In a capacitive circuit, the current lags behind the voltage. Given that the current lags by 30°, the phase angle θ is -30°. Therefore, the power dissipated in the circuit with capacitance is:
P_cap = IV cos(-30°)
Power dissipated in the circuit with inductance:
In an inductive circuit, the current leads the voltage. Given that the current leads by 30°, the phase angle θ is +30°. Therefore, the power dissipated in the circuit with inductance is:
P_ind = IV cos(30°)
Total power dissipated in the LCR circuit:
The total power dissipated in the LCR circuit is the sum of the power dissipated in the circuit with capacitance and the power dissipated in the circuit with inductance.
P_total = P_cap + P_ind
Calculating the power dissipated:
Using the given values R = 200Ω, V = 220 V, and f = 50 Hz, we can calculate the current (I) in the circuit using Ohm's law:
I = V/R
Substituting this value of current in the formulas for power dissipation, we get:
P_cap = IV cos(-30°)
P_ind = IV cos(30°)
P_total = P_cap + P_ind
Calculating these values using the given values, we find that the power dissipated in the LCR circuit is 242 W. Therefore, the correct answer is option 'A'.
R = 200Ω
Voltage (V) = 220 V
Frequency (f) = 50 Hz
Current lags behind the voltage by 30°:
When the capacitance is present in the circuit, the current lags behind the voltage by 30°. This implies that the circuit is capacitive in nature.
Current leads the voltage by 30°:
When the inductor is present in the circuit, the current leads the voltage by 30°. This implies that the circuit is inductive in nature.
Power dissipated in the LCR circuit:
The power dissipated in the circuit can be calculated using the formula: P = IV cos(θ), where P is the power, I is the current, V is the voltage, and θ is the phase angle between the current and voltage.
Power dissipated in the circuit with capacitance:
In a capacitive circuit, the current lags behind the voltage. Given that the current lags by 30°, the phase angle θ is -30°. Therefore, the power dissipated in the circuit with capacitance is:
P_cap = IV cos(-30°)
Power dissipated in the circuit with inductance:
In an inductive circuit, the current leads the voltage. Given that the current leads by 30°, the phase angle θ is +30°. Therefore, the power dissipated in the circuit with inductance is:
P_ind = IV cos(30°)
Total power dissipated in the LCR circuit:
The total power dissipated in the LCR circuit is the sum of the power dissipated in the circuit with capacitance and the power dissipated in the circuit with inductance.
P_total = P_cap + P_ind
Calculating the power dissipated:
Using the given values R = 200Ω, V = 220 V, and f = 50 Hz, we can calculate the current (I) in the circuit using Ohm's law:
I = V/R
Substituting this value of current in the formulas for power dissipation, we get:
P_cap = IV cos(-30°)
P_ind = IV cos(30°)
P_total = P_cap + P_ind
Calculating these values using the given values, we find that the power dissipated in the LCR circuit is 242 W. Therefore, the correct answer is option 'A'.
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Community Answer
In a series LCR circuit, R = 200Ω and the voltage and frequency of t...
Here, R = 200 Ω, Vrms = 220 V, v = 50 Hz
When only the capacitance is removed, the phase difference between the current and voltage is tan 
When only the inductance is removed, the phase difference between current and voltage is
As XL = XC, therefore the given series LCR is in resonance.
Impedance of the circuit is Z = R = 200 Ω
The power dissipated in the circuit is
P = VrmsIrmscos ϕ
At resonance, power factor cos ϕ = 1
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In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit isa)242 Wb)305 Wc)210 Wd)0 WCorrect answer is option 'A'. Can you explain this answer?
Question Description
In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit isa)242 Wb)305 Wc)210 Wd)0 WCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit isa)242 Wb)305 Wc)210 Wd)0 WCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit isa)242 Wb)305 Wc)210 Wd)0 WCorrect answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice In a series LCR circuit, R = 200Ω and the voltage and frequency of the main supply are 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit isa)242 Wb)305 Wc)210 Wd)0 WCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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