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In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit in
capacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]
sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 J
in the circuit is given as [JEE (Main)-2020]?
capacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]
sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 J
in the circuit is given as [JEE (Main)-2020]?
Most Upvoted Answer
In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipate...
Inductance and Capacitance in LC Circuit
The given parameters for the LC circuit are:
- Inductance (L) = 40 mH = 0.04 H
- Capacitance (C) = 100 F
- Frequency (f) = 50 Hz
- Voltage (V(t)) = 10 sin(314t)
Energy in the Circuit
The energy stored in a capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
For the given circuit, substituting the values:
- \( C = 100 \, F \)
- \( V = 10 \, V \)
Calculating the energy:
\[
U = \frac{1}{2} \times 100 \times (10)^2
\]
\[
U = \frac{1}{2} \times 100 \times 100 = 5000 \, J
\]
Current Calculation
The current \( I(t) \) in the circuit can be expressed as:
\[
I(t) = \frac{V(t)}{Z}
\]
Where \( Z \) is the impedance of the circuit. The impedance in an LC circuit is given by:
\[
Z = \sqrt{L/C}
\]
Calculating \( Z \):
\[
Z = \sqrt{\frac{0.04}{100}} = \sqrt{0.0004} = 0.02 \, \Omega
\]
Now, substituting into the current equation:
\[
I(t) = \frac{10 \sin(314t)}{0.02} = 500 \sin(314t) \, A
\]
Final Energy Dissipation
To find the energy dissipated in the circuit, use the average power formula over one cycle:
\[
P_{avg} = \frac{V_{rms}^2}{Z}
\]
Calculating \( V_{rms} \):
\[
V_{rms} = \frac{10}{\sqrt{2}}
\]
Thus,
\[
P_{avg} = \frac{(10/\sqrt{2})^2}{0.02} = \frac{50}{0.02} = 2500 \, W
\]
This results in the total energy dissipated in one cycle over a time period derived from the frequency.
Overall, the energy dissipated in the circuit is significant, confirming the calculations align with the principles of an LC circuit in sinusoidal voltage conditions.
The given parameters for the LC circuit are:
- Inductance (L) = 40 mH = 0.04 H
- Capacitance (C) = 100 F
- Frequency (f) = 50 Hz
- Voltage (V(t)) = 10 sin(314t)
Energy in the Circuit
The energy stored in a capacitor is given by the formula:
\[
U = \frac{1}{2} C V^2
\]
For the given circuit, substituting the values:
- \( C = 100 \, F \)
- \( V = 10 \, V \)
Calculating the energy:
\[
U = \frac{1}{2} \times 100 \times (10)^2
\]
\[
U = \frac{1}{2} \times 100 \times 100 = 5000 \, J
\]
Current Calculation
The current \( I(t) \) in the circuit can be expressed as:
\[
I(t) = \frac{V(t)}{Z}
\]
Where \( Z \) is the impedance of the circuit. The impedance in an LC circuit is given by:
\[
Z = \sqrt{L/C}
\]
Calculating \( Z \):
\[
Z = \sqrt{\frac{0.04}{100}} = \sqrt{0.0004} = 0.02 \, \Omega
\]
Now, substituting into the current equation:
\[
I(t) = \frac{10 \sin(314t)}{0.02} = 500 \sin(314t) \, A
\]
Final Energy Dissipation
To find the energy dissipated in the circuit, use the average power formula over one cycle:
\[
P_{avg} = \frac{V_{rms}^2}{Z}
\]
Calculating \( V_{rms} \):
\[
V_{rms} = \frac{10}{\sqrt{2}}
\]
Thus,
\[
P_{avg} = \frac{(10/\sqrt{2})^2}{0.02} = \frac{50}{0.02} = 2500 \, W
\]
This results in the total energy dissipated in one cycle over a time period derived from the frequency.
Overall, the energy dissipated in the circuit is significant, confirming the calculations align with the principles of an LC circuit in sinusoidal voltage conditions.
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In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]?
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In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]?.
In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In LC circuit the inductance L = 40 mH and 50 Hz. The energy dissipated in the circuit incapacitance C = 100 F.If a voltage Vt) = 10 60 s is [JEE (Main)-2019]sin (314 t) is applied to the circuit, the current (1) 5.65 × 102 J (2) 5.17 × 102 Jin the circuit is given as [JEE (Main)-2020]?.
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