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A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803
H and a capacitance of 10 uF all in series. The Time (heat capacity
When an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume no
applied to the coil, the applied voltage leads
the current by 45°. The self-inductance of the loss of heat to the surrounding) is close to?
H and a capacitance of 10 uF all in series. The Time (heat capacity
When an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume no
applied to the coil, the applied voltage leads
the current by 45°. The self-inductance of the loss of heat to the surrounding) is close to?
Most Upvoted Answer
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an ...
Given Data
- Frequency (f): 1000 Hz
- Voltage (V_rms): 20 V
- Resistance (R): 100 Ω
- Inductance (L): 0.1803 H
- Capacitance (C): 10 µF
- Temperature rise (ΔT): 10°C
- Heat capacity (C_Heat): 2 J/°C
Impedance Calculation
- The impedance (Z) in an RLC circuit is given by:
Z = sqrt(R^2 + (X_L - X_C)^2)
- Where:
- X_L = 2πfL
- X_C = 1/(2πfC)
- Calculate X_L and X_C:
X_L = 2π(1000)(0.1803) ≈ 1136.8 Ω
X_C = 1/(2π(1000)(10 x 10^-6)) ≈ 15.9 Ω
- Therefore,
Z = sqrt(100^2 + (1136.8 - 15.9)^2) ≈ 1136.8 Ω
Current Calculation
- The rms current (I_rms) is calculated as:
I_rms = V_rms / Z
I_rms = 20 V / 1136.8 Ω ≈ 0.0176 A
Power Dissipation and Heat Generation
- The power dissipated (P) in the circuit is:
P = I_rms^2 * R
P = (0.0176)^2 * 100 ≈ 0.031 J/s
Heating Calculation
- To find the time (t) for a 10°C rise:
Energy = C_Heat * ΔT = 2 J/°C * 10°C = 20 J
- Time can be calculated as:
t = Energy / Power = 20 J / 0.031 J/s ≈ 645.16 s
Conclusion
- The time required to heat the coil by 10°C is approximately 645.16 seconds under the given conditions.
- Frequency (f): 1000 Hz
- Voltage (V_rms): 20 V
- Resistance (R): 100 Ω
- Inductance (L): 0.1803 H
- Capacitance (C): 10 µF
- Temperature rise (ΔT): 10°C
- Heat capacity (C_Heat): 2 J/°C
Impedance Calculation
- The impedance (Z) in an RLC circuit is given by:
Z = sqrt(R^2 + (X_L - X_C)^2)
- Where:
- X_L = 2πfL
- X_C = 1/(2πfC)
- Calculate X_L and X_C:
X_L = 2π(1000)(0.1803) ≈ 1136.8 Ω
X_C = 1/(2π(1000)(10 x 10^-6)) ≈ 15.9 Ω
- Therefore,
Z = sqrt(100^2 + (1136.8 - 15.9)^2) ≈ 1136.8 Ω
Current Calculation
- The rms current (I_rms) is calculated as:
I_rms = V_rms / Z
I_rms = 20 V / 1136.8 Ω ≈ 0.0176 A
Power Dissipation and Heat Generation
- The power dissipated (P) in the circuit is:
P = I_rms^2 * R
P = (0.0176)^2 * 100 ≈ 0.031 J/s
Heating Calculation
- To find the time (t) for a 10°C rise:
Energy = C_Heat * ΔT = 2 J/°C * 10°C = 20 J
- Time can be calculated as:
t = Energy / Power = 20 J / 0.031 J/s ≈ 645.16 s
Conclusion
- The time required to heat the coil by 10°C is approximately 645.16 seconds under the given conditions.
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A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to?
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A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to?.
A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 750 Hz, 20 V (rms) source is connected to a resistance of 100 s, an inductance of 0.1803H and a capacitance of 10 uF all in series. The Time (heat capacityWhen an AC signal of frequency 1000 Hz is 2 J/°C) will get heated by 10°C. (assume noapplied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close to?.
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