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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?
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.
Free Test
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
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