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A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-
- a)(loge (2))
- b)(10-3loge (2))
- c)(25loge (2))
- d)(1/40loge (2))
Correct answer is option 'D'. Can you explain this answer?
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A solenoid of inductance 250 mH and resistance is connected to a batt...
Let at time t; magnetic energy reaches to (1/4) of its maximum.
Energy, E = 1/2Li2
i.e., current that time should be (max/2)
1/2 = e-tR/L
Most Upvoted Answer
A solenoid of inductance 250 mH and resistance is connected to a batt...
Given data:
Inductance of solenoid (L) = 250 mH = 0.25 H
Resistance of solenoid (R) = ?
Time taken for magnetic energy to reach (1/4th) of its maximum value = ?
We know that the magnetic energy stored in an inductor is given by the formula:
E = (1/2) * L * I^2
where E is the magnetic energy, L is the inductance, and I is the current flowing through the inductor.
To calculate the time taken for the magnetic energy to reach (1/4th) of its maximum value, we need to find the current at that time.
Let's assume that the current at time t is I(t), and the maximum current is I(max).
Key Point 1: Relationship between current and time in an RL circuit
In an RL circuit, the current through the inductor increases with time and approaches a maximum value. The rate of change of current with time is given by the equation:
dI/dt = (V - IR) / L
where V is the applied voltage, R is the resistance, and L is the inductance.
Key Point 2: Energy dissipated in the resistor
The energy dissipated in the resistor is given by the formula:
E_R = (1/2) * R * I^2
where E_R is the energy dissipated, R is the resistance, and I is the current flowing through the resistor.
Key Point 3: Relationship between energy stored in the inductor and energy dissipated in the resistor
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor. Therefore, we have:
dE_R/dt = -dE/dt
where dE_R/dt is the rate of energy dissipation in the resistor and dE/dt is the rate of change of energy stored in the inductor.
Key Point 4: Substituting the formulas
Substituting the formulas for energy in the above equation, we get:
(1/2) * R * I^2 = -dE/dt
Rearranging the equation, we get:
dE/dt = -(1/2) * R * I^2
Key Point 5: Solving the differential equation
We can solve the above differential equation to find the current as a function of time.
dE/dt = -(1/2) * R * I^2
=> dE/E = -(1/2) * R * I^2 * dt
Integrating both sides, we get:
∫dE/E = ∫-(1/2) * R * I^2 * dt
=> ln(E) = -(1/2) * R * ∫I^2 * dt
Key Point 6: Calculating the time taken for magnetic energy to reach (1/4th) of its maximum value
Now, we need to calculate the time taken for the magnetic energy to reach (1/4th) of its maximum value, which is equivalent to finding the time when E = (1/4) * E(max).
Let's assume that the current at this time is I(1/4), and the maximum current is I(max).
Substituting the values in the equation obtained from integrating
Inductance of solenoid (L) = 250 mH = 0.25 H
Resistance of solenoid (R) = ?
Time taken for magnetic energy to reach (1/4th) of its maximum value = ?
We know that the magnetic energy stored in an inductor is given by the formula:
E = (1/2) * L * I^2
where E is the magnetic energy, L is the inductance, and I is the current flowing through the inductor.
To calculate the time taken for the magnetic energy to reach (1/4th) of its maximum value, we need to find the current at that time.
Let's assume that the current at time t is I(t), and the maximum current is I(max).
Key Point 1: Relationship between current and time in an RL circuit
In an RL circuit, the current through the inductor increases with time and approaches a maximum value. The rate of change of current with time is given by the equation:
dI/dt = (V - IR) / L
where V is the applied voltage, R is the resistance, and L is the inductance.
Key Point 2: Energy dissipated in the resistor
The energy dissipated in the resistor is given by the formula:
E_R = (1/2) * R * I^2
where E_R is the energy dissipated, R is the resistance, and I is the current flowing through the resistor.
Key Point 3: Relationship between energy stored in the inductor and energy dissipated in the resistor
The rate at which energy is dissipated in the resistor is equal to the rate at which energy is stored in the inductor. Therefore, we have:
dE_R/dt = -dE/dt
where dE_R/dt is the rate of energy dissipation in the resistor and dE/dt is the rate of change of energy stored in the inductor.
Key Point 4: Substituting the formulas
Substituting the formulas for energy in the above equation, we get:
(1/2) * R * I^2 = -dE/dt
Rearranging the equation, we get:
dE/dt = -(1/2) * R * I^2
Key Point 5: Solving the differential equation
We can solve the above differential equation to find the current as a function of time.
dE/dt = -(1/2) * R * I^2
=> dE/E = -(1/2) * R * I^2 * dt
Integrating both sides, we get:
∫dE/E = ∫-(1/2) * R * I^2 * dt
=> ln(E) = -(1/2) * R * ∫I^2 * dt
Key Point 6: Calculating the time taken for magnetic energy to reach (1/4th) of its maximum value
Now, we need to calculate the time taken for the magnetic energy to reach (1/4th) of its maximum value, which is equivalent to finding the time when E = (1/4) * E(max).
Let's assume that the current at this time is I(1/4), and the maximum current is I(max).
Substituting the values in the equation obtained from integrating
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A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer?
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A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer?.
A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solenoid of inductance 250 mH and resistance is connected to a battery. The time taken for the magnetic energy to reach (1/4th) of its maximum value is-a)(loge (2))b)(10-3loge (2))c)(25loge (2))d)(1/40loge (2))Correct answer is option 'D'. Can you explain this answer?.
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