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A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum are
- a)2.93 A, 1.55 ms
- b)2.53 A, 1.55 ms
- c)3.03 A, 1.50 ms
- d)3.23 A, 1.55 ms
Correct answer is option 'D'. Can you explain this answer?
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To find the maximum current in the circuit and the time lag between the current maximum and the voltage maximum, we need to analyze the behavior of the capacitor in a series circuit.
1. Calculating the impedance of the circuit:
- The impedance of the capacitor in series is given by Zc = 1/(jωC), where j is the imaginary unit, ω is the angular frequency (2πf), and C is the capacitance.
- Plugging in the values: Zc = 1/(j * 2π * 60 * 100 * 10^(-6)) = -j265.26 Ω
2. Calculating the total impedance of the circuit:
- The total impedance (Zt) of the circuit is given by Zt = R + Zc, where R is the resistance.
- Plugging in the values: Zt = 40 + (-j265.26) = 40 - j265.26 Ω
3. Calculating the current in the circuit:
- The maximum current (Imax) in the circuit is given by Imax = Vmax / |Zt|, where Vmax is the maximum voltage of the supply and |Zt| is the magnitude of the total impedance.
- Plugging in the values: Imax = 110 / |40 - j265.26| = 3.23 A
4. Calculating the time lag between current and voltage maxima:
- The time lag (Δt) between the current maximum and the voltage maximum is given by Δt = φ / (2πf), where φ is the phase angle between the current and the voltage.
- The phase angle (φ) can be calculated using the arctan function: φ = arctan(Im / Re), where Im is the imaginary part of Zt and Re is the real part of Zt.
- Plugging in the values: φ = arctan(-265.26 / 40) = -1.55 radians
- Plugging in φ into the equation for Δt: Δt = -1.55 / (2π * 60) = 1.55 ms
Therefore, the correct answer is option D: 3.23 A, 1.55 ms.
1. Calculating the impedance of the circuit:
- The impedance of the capacitor in series is given by Zc = 1/(jωC), where j is the imaginary unit, ω is the angular frequency (2πf), and C is the capacitance.
- Plugging in the values: Zc = 1/(j * 2π * 60 * 100 * 10^(-6)) = -j265.26 Ω
2. Calculating the total impedance of the circuit:
- The total impedance (Zt) of the circuit is given by Zt = R + Zc, where R is the resistance.
- Plugging in the values: Zt = 40 + (-j265.26) = 40 - j265.26 Ω
3. Calculating the current in the circuit:
- The maximum current (Imax) in the circuit is given by Imax = Vmax / |Zt|, where Vmax is the maximum voltage of the supply and |Zt| is the magnitude of the total impedance.
- Plugging in the values: Imax = 110 / |40 - j265.26| = 3.23 A
4. Calculating the time lag between current and voltage maxima:
- The time lag (Δt) between the current maximum and the voltage maximum is given by Δt = φ / (2πf), where φ is the phase angle between the current and the voltage.
- The phase angle (φ) can be calculated using the arctan function: φ = arctan(Im / Re), where Im is the imaginary part of Zt and Re is the real part of Zt.
- Plugging in the values: φ = arctan(-265.26 / 40) = -1.55 radians
- Plugging in φ into the equation for Δt: Δt = -1.55 / (2π * 60) = 1.55 ms
Therefore, the correct answer is option D: 3.23 A, 1.55 ms.
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A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer?
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A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer?.
A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A 100 μF capacitor in series with a 40Ωresistance is connected to a 110 V, 60 Hz supply. Maximum current in the circuit and time lag between the current maximum and the voltage maximum area)2.93 A, 1.55 msb)2.53 A, 1.55 msc)3.03 A, 1.50 msd)3.23 A, 1.55 msCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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