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A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and the resultant current will be:
- a)tan-1 (0.17)
- b)tan-1 (9.46)
- c)tan-1 (0.30)
- d)tan-1 (13.33)
Correct answer is option 'A'. Can you explain this answer?
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A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series...
ΜF, and R = 50 Ω. Find the maximum current through the circuit and the phase angle between the current and the applied voltage.
To find the maximum current through the circuit, we can use Ohm's Law. The impedance of the circuit is given by Z = sqrt(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance.
XL = 2πfL = 2π(3000)(0.01) = 188.5 Ω
XC = 1/(2πfC) = 1/(2π(3000)(25 * 10^-6)) = 106.1 Ω
Z = sqrt((50)^2 + (188.5 - 106.1)^2) = 164.9 Ω
The maximum current (Imax) is given by Imax = Vmax / Z, where Vmax is the maximum voltage.
Vmax = 210 V
Imax = 210 / 164.9 = 1.27 A
To find the phase angle between the current and the applied voltage, we can use the formula tan(θ) = (XL - XC) / R.
tan(θ) = (188.5 - 106.1) / 50 = 1.645
Taking the arctan of both sides, we find θ = arctan(1.645) = 57.71 degrees.
Therefore, the maximum current through the circuit is 1.27 A and the phase angle between the current and the applied voltage is 57.71 degrees.
To find the maximum current through the circuit, we can use Ohm's Law. The impedance of the circuit is given by Z = sqrt(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance.
XL = 2πfL = 2π(3000)(0.01) = 188.5 Ω
XC = 1/(2πfC) = 1/(2π(3000)(25 * 10^-6)) = 106.1 Ω
Z = sqrt((50)^2 + (188.5 - 106.1)^2) = 164.9 Ω
The maximum current (Imax) is given by Imax = Vmax / Z, where Vmax is the maximum voltage.
Vmax = 210 V
Imax = 210 / 164.9 = 1.27 A
To find the phase angle between the current and the applied voltage, we can use the formula tan(θ) = (XL - XC) / R.
tan(θ) = (188.5 - 106.1) / 50 = 1.645
Taking the arctan of both sides, we find θ = arctan(1.645) = 57.71 degrees.
Therefore, the maximum current through the circuit is 1.27 A and the phase angle between the current and the applied voltage is 57.71 degrees.
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Community Answer
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series...
XL = 3000 × 10 × 10-3 = 30 Ω
So, θ = tan-1 (0.17)
So, θ = tan-1 (0.17)
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A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and the resultant current will be:a)tan-1 (0.17)b)tan-1 (9.46)c)tan-1 (0.30)d)tan-1 (13.33)Correct answer is option 'A'. Can you explain this answer?
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A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and the resultant current will be:a)tan-1 (0.17)b)tan-1 (9.46)c)tan-1 (0.30)d)tan-1 (13.33)Correct 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 A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and the resultant current will be:a)tan-1 (0.17)b)tan-1 (9.46)c)tan-1 (0.30)d)tan-1 (13.33)Correct 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 A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and the resultant current will be:a)tan-1 (0.17)b)tan-1 (9.46)c)tan-1 (0.30)d)tan-1 (13.33)Correct answer is option 'A'. Can you explain this answer?.
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