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When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid are
- a), 0.55 henry
- b), 0.86 henry
- c), 1.0 henry
- d), 0.93
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flow...
Explanation:
Given,
DC voltage, VDC = 100 V
Current in DC, IDC = 1.0 A
AC voltage, VAC = 100 V
Current in AC, IAC = 0.5 A
Frequency of AC, f = 50 Hz
Finding the Impedance:
Impedance of the solenoid can be found using Ohm's law as follows:
Z = V / I
For DC, ZDC = VDC / IDC = 100 / 1.0 = 100 Ω
For AC, ZAC = VAC / IAC = 100 / 0.5 = 200 Ω
Finding the Inductance:
Impedance of a solenoid is given by the following formula:
Z = √(R2 + XL2)
where R is the resistance of the solenoid and XL is the inductive reactance of the solenoid.
Resistance of the solenoid can be calculated as follows:
R = V / I = 100 / 1.0 = 100 Ω
Finding the Inductive Reactance:
Inductive reactance can be calculated using the following formula:
XL = ZAC / √2 - R
where √2 is the root of 2 and R is the resistance of the solenoid.
Substituting the values, we get:
XL = 200 / √2 - 100
XL = 200 / 0.414 - 100
XL = 200 / 0.414 - 100
XL = 200 / 0.414 - 100
XL = 0.586 * 200 - 100
XL = 117.2 - 100
XL = 17.2 Ω
Finding the Inductance:
Inductance can be calculated using the following formula:
XL = 2πfL
where f is the frequency of AC and L is the inductance of the solenoid.
Substituting the values, we get:
17.2 = 2π * 50 * L
L = 17.2 / (2π * 50)
L = 0.055 henry
Final Answer:
Therefore, the impedance and inductance of the solenoid are
Given,
DC voltage, VDC = 100 V
Current in DC, IDC = 1.0 A
AC voltage, VAC = 100 V
Current in AC, IAC = 0.5 A
Frequency of AC, f = 50 Hz
Finding the Impedance:
Impedance of the solenoid can be found using Ohm's law as follows:
Z = V / I
For DC, ZDC = VDC / IDC = 100 / 1.0 = 100 Ω
For AC, ZAC = VAC / IAC = 100 / 0.5 = 200 Ω
Finding the Inductance:
Impedance of a solenoid is given by the following formula:
Z = √(R2 + XL2)
where R is the resistance of the solenoid and XL is the inductive reactance of the solenoid.
Resistance of the solenoid can be calculated as follows:
R = V / I = 100 / 1.0 = 100 Ω
Finding the Inductive Reactance:
Inductive reactance can be calculated using the following formula:
XL = ZAC / √2 - R
where √2 is the root of 2 and R is the resistance of the solenoid.
Substituting the values, we get:
XL = 200 / √2 - 100
XL = 200 / 0.414 - 100
XL = 200 / 0.414 - 100
XL = 200 / 0.414 - 100
XL = 0.586 * 200 - 100
XL = 117.2 - 100
XL = 17.2 Ω
Finding the Inductance:
Inductance can be calculated using the following formula:
XL = 2πfL
where f is the frequency of AC and L is the inductance of the solenoid.
Substituting the values, we get:
17.2 = 2π * 50 * L
L = 17.2 / (2π * 50)
L = 0.055 henry
Final Answer:
Therefore, the impedance and inductance of the solenoid are
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When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 Correct answer is option 'A'. Can you explain this answer?
Question Description
When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 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 When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 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 When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 Correct answer is option 'A'. Can you explain this answer?.
When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 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 When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 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 When 100 V D.C. is supplied across solenoid, a current of 1.0 amp flows in it. When 100 V A.C. is applied across the same coil, the current drops to 0.5A. If the frequency of A.C. source is 50Hz, then the impedance and inductance of the solenoid area) 200 Ω , 0.55 henry b) 100 Ω , 0.86 henry c) 200 Ω , 1.0 henry d) 100 Ω , 0.93 Correct answer is option 'A'. Can you explain this answer?.
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