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When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry?
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When 100 volts DC is applied across a solenoid, a current of 1.0 amper...
Correct option (a) 0.55 H
Explanation:
For dc, V = IR (R is resistance of inductor)
Explanation:
For dc, V = IR (R is resistance of inductor)
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When 100 volts DC is applied across a solenoid, a current of 1.0 amper...
Impedance and Inductance of a Solenoid
To determine the impedance and inductance of the solenoid, we need to analyze the behavior of the solenoid under DC and AC conditions.
DC Condition:
When 100 volts DC is applied across the solenoid, a current of 1.0 ampere flows through it.
AC Condition:
When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. The frequency of the AC source is given as 50 Hz.
Impedance (Z):
Impedance is the opposition offered by an electrical circuit to the flow of alternating current. It is a complex quantity that consists of both resistance and reactance. In the case of a solenoid, the reactance is due to its inductance.
The formula for impedance is given by:
Z = V/I
Where,
Z is the impedance
V is the voltage applied
I is the current flowing through the circuit
For the DC condition:
Z_DC = 100V / 1.0A
Z_DC = 100 ohms
For the AC condition:
Z_AC = 100V / 0.5A
Z_AC = 200 ohms
Inductance (L):
Inductance is a property of an electrical circuit that opposes any change in the current flowing through it. It is measured in henries (H).
The formula for inductive reactance is given by:
XL = 2πfL
Where,
XL is the inductive reactance
π is a mathematical constant approximately equal to 3.14159
f is the frequency of the AC source
L is the inductance of the solenoid
For the AC condition:
XL = 2π(50Hz)L
XL = 100πL
From the given information, we know that the current drops to half when AC is applied. This indicates that the inductive reactance has doubled. Therefore, we can equate the inductive reactance for the DC and AC conditions.
For the DC condition:
XL_DC = 0 (since it's a DC condition)
For the AC condition:
XL_AC = 2 × XL_DC
100πL = 2 × 0
100πL = 0
Since 100πL = 0, it implies that L = 0.
Therefore, the inductance of the solenoid is 0 henries.
Conclusion:
Based on the given information, the impedance of the solenoid is 200 ohms and the inductance is 0 henries.
The correct answer is Option A) 200 ohms and 0.55 henry.
To determine the impedance and inductance of the solenoid, we need to analyze the behavior of the solenoid under DC and AC conditions.
DC Condition:
When 100 volts DC is applied across the solenoid, a current of 1.0 ampere flows through it.
AC Condition:
When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. The frequency of the AC source is given as 50 Hz.
Impedance (Z):
Impedance is the opposition offered by an electrical circuit to the flow of alternating current. It is a complex quantity that consists of both resistance and reactance. In the case of a solenoid, the reactance is due to its inductance.
The formula for impedance is given by:
Z = V/I
Where,
Z is the impedance
V is the voltage applied
I is the current flowing through the circuit
For the DC condition:
Z_DC = 100V / 1.0A
Z_DC = 100 ohms
For the AC condition:
Z_AC = 100V / 0.5A
Z_AC = 200 ohms
Inductance (L):
Inductance is a property of an electrical circuit that opposes any change in the current flowing through it. It is measured in henries (H).
The formula for inductive reactance is given by:
XL = 2πfL
Where,
XL is the inductive reactance
π is a mathematical constant approximately equal to 3.14159
f is the frequency of the AC source
L is the inductance of the solenoid
For the AC condition:
XL = 2π(50Hz)L
XL = 100πL
From the given information, we know that the current drops to half when AC is applied. This indicates that the inductive reactance has doubled. Therefore, we can equate the inductive reactance for the DC and AC conditions.
For the DC condition:
XL_DC = 0 (since it's a DC condition)
For the AC condition:
XL_AC = 2 × XL_DC
100πL = 2 × 0
100πL = 0
Since 100πL = 0, it implies that L = 0.
Therefore, the inductance of the solenoid is 0 henries.
Conclusion:
Based on the given information, the impedance of the solenoid is 200 ohms and the inductance is 0 henries.
The correct answer is Option A) 200 ohms and 0.55 henry.
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When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry?
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When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry? 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 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry?.
When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry? 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 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry?.
Solutions for When 100 volts DC is applied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts AC is applied across the same coil, the current drops to 0.5 ampere. If the frequency of AC source is 50 Hz, then the impedance and inductance of solenoid are: A) 200 ohm and 0.55 henry B) 100 ohm and 0.86 henry C) 200 ohm and 1.0 henry D) 100 ohm and 0.93 henry? in English & in Hindi are available as part of our courses for JEE.
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