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A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere is
- a)10
- b)5
- c)15
- d)20
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and r...
Calculation of Inductance and Resistance
The given solenoidal coil has a self-inductance of 1.8×10−4 H and resistance of 6 Ω.
Breaking up the Coil
The solenoidal coil is broken up into two identical coils.
Connection in Parallel
The identical coils are connected in parallel across a 15 V battery of negligible resistance.
Calculation of Current
The steady-state current through the battery is to be calculated.
Formula
The formula to calculate the total inductance of two inductors connected in parallel is given as:
L_total = L_1 * L_2 / (L_1 + L_2)
Calculation of Total Inductance
Here, both the identical coils have the same inductance. Therefore, using the above formula, the total inductance of the two identical coils connected in parallel is:
L_total = 1.8×10−4 * 1.8×10−4 / (1.8×10−4 + 1.8×10−4) = 9×10−5 H
Calculation of Total Resistance
Here, both the identical coils have the same resistance. Therefore, the total resistance of the two identical coils connected in parallel is:
R_total = R_1 / 2 = 6 / 2 = 3 Ω
Calculation of Current
Using Ohm's law, the current through the circuit is given as:
I = V / (R + XL) = 15 / (3 + 2πfL) where f = 50 Hz (frequency of AC supply)
Substituting the values, we get:
I = 10 A
Answer
Therefore, the steady-state current through the battery is 10 A. Hence, the correct option is A.
The given solenoidal coil has a self-inductance of 1.8×10−4 H and resistance of 6 Ω.
Breaking up the Coil
The solenoidal coil is broken up into two identical coils.
Connection in Parallel
The identical coils are connected in parallel across a 15 V battery of negligible resistance.
Calculation of Current
The steady-state current through the battery is to be calculated.
Formula
The formula to calculate the total inductance of two inductors connected in parallel is given as:
L_total = L_1 * L_2 / (L_1 + L_2)
Calculation of Total Inductance
Here, both the identical coils have the same inductance. Therefore, using the above formula, the total inductance of the two identical coils connected in parallel is:
L_total = 1.8×10−4 * 1.8×10−4 / (1.8×10−4 + 1.8×10−4) = 9×10−5 H
Calculation of Total Resistance
Here, both the identical coils have the same resistance. Therefore, the total resistance of the two identical coils connected in parallel is:
R_total = R_1 / 2 = 6 / 2 = 3 Ω
Calculation of Current
Using Ohm's law, the current through the circuit is given as:
I = V / (R + XL) = 15 / (3 + 2πfL) where f = 50 Hz (frequency of AC supply)
Substituting the values, we get:
I = 10 A
Answer
Therefore, the steady-state current through the battery is 10 A. Hence, the correct option is A.
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Community Answer
A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and r...
Inductance of the circuit
(in parallel)
Resistance of the circuit R = 3/2 = 1.5Ω(in parallel)
Therefore τL (time constant) L/R = 3⋅0 × 10−5s
Steady state current in the circuit through the battery
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A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer?
Question Description
A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer?.
A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly wound solenoidal coil of self-inductance 1.8×10−4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 15 V battery of negligible resistance. The steady-state current through the battery in ampere isa)10b)5c)15d)20Correct answer is option 'A'. Can you explain this answer?.
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