Sure, let's solve this step by step.

(A) The time elapsed before the current acquires half of its steady – state value.

The current in an LR circuit at any time 't' after the switch is closed is given by the formula:

I = E/R * (1 - e^(-Rt/L))

where:
E = emf of the battery
R = resistance
L = inductance
t = time

We need to find the time when the current is half of its steady state value. The steady state value of the current (I) is E/R. So, we need to solve for t in the following equation:

0.5 * E/R = E/R * (1 - e^(-Rt/L))

Solving this equation gives:

0.5 = 1 - e^(-Rt/L)

Rearranging terms gives:

e^(-Rt/L) = 0.5

Taking the natural logarithm of both sides gives:

-Rt/L = ln(0.5)

Solving for t gives:

t = -L/R * ln(0.5)

Substituting the given values L = 1H, R = 100Ω, and ln(0.5) = -0.693, we get:

t = -1/100 * -0.693 = 0.00693 s

So, the time elapsed before the current acquires half of its steady state value is approximately 0.00693 seconds.

(B) The energy stored in the magnetic field associated with the coil at instant 15 m s after the circuit is switched on.

The energy (W) stored in the magnetic field of an inductor is given by the formula:

W = 0.5 * L * I^2

We already know that the current at any time 't' is given by:

I = E/R * (1 - e^(-Rt/L))

Substituting the given values E = 6V, R = 100Ω, L = 1H, and t = 15ms = 0.015s, we get:

I = 6/100 * (1 - e^(-100*0.015))

I = 0.06 * (1 - e^-1.5)

Given that e^-1.5 = 0.25, we get:

I = 0.06 * (1 - 0.25) = 0.06 * 0.75 = 0.045 A

Substituting I = 0.045A and L = 1H into the formula for W gives:

W = 0.5 * 1 * (0.045)^2 = 0.0010125 J

So, the energy stored in the magnetic field associated with the coil at instant 15 ms after the circuit is switched on is approximately 0.0010125 Joules.

Question

Sure, let's solve this step by step.

(A) The time elapsed before the current acquires half of its steady – state value.

The current in an LR circuit at any time 't' after the switch is closed is given by the formula:

I = E/R * (1 - e^(-Rt/L))

where:
E = emf of the battery
R = resistance
L = inductance
t = time

We need to find the time when the current is half of its steady state value. The steady state value of the current (I) is E/R. So, we need to solve for t in the following equation:

0.5 * E/R = E/R * (1 - e^(-Rt/L))

Solving this equation gives:

0.5 = 1 - e^(-Rt/L)

Rearranging terms gives:

e^(-Rt/L) = 0.5

Taking the natural logarithm of both sides gives:

-Rt/L = ln(0.5)

Solving for t gives:

t = -L/R * ln(0.5)

Substituting the given values L = 1H, R = 100Ω, and ln(0.5) = -0.693, we get:

t = -1/100 * -0.693 = 0.00693 s

So, the time elapsed before the current acquires half of its steady state value is approximately 0.00693 seconds.

(B) The energy stored in the magnetic field associated with the coil at instant 15 m s after the circuit is switched on.

The energy (W) stored in the magnetic field of an inductor is given by the formula:

W = 0.5 * L * I^2

We already know that the current at any time 't' is given by:

I = E/R * (1 - e^(-Rt/L))

Substituting the given values E = 6V, R = 100Ω, L = 1H, and t = 15ms = 0.015s, we get:

I = 6/100 * (1 - e^(-100*0.015))

I = 0.06 * (1 - e^-1.5)

Given that e^-1.5 = 0.25, we get:

I = 0.06 * (1 - 0.25) = 0.06 * 0.75 = 0.045 A

Substituting I = 0.045A and L = 1H into the formula for W gives:

W = 0.5 * 1 * (0.045)^2 = 0.0010125 J

So, the energy stored in the magnetic field associated with the coil at instant 15 ms after the circuit is switched on is approximately 0.0010125 Joules.

Knowee AI · Accepted Answer

Sure, let's solve this step by step.

(A) The time elapsed before the current acquires half of its steady – state value.

The current in an LR circuit at any time 't' after the switch is closed is given by the formula:

I = E/R * (1 - e^(-Rt/L))

where:
E = emf of the battery
R = resistance
L = inductance
t = time

We need to find the time when the current is half of its steady state value. The steady state value of the current (I) is E/R. So, we need to solve for t in the following equation:

0.5 * E/R = E/R * (1 - e^(-Rt/L))

Solving this equation gives:

0.5 = 1 - e^(-Rt/L)

Rearranging terms gives:

e^(-Rt/L) = 0.5

Taking the natural logarithm of both sides gives:

-Rt/L = ln(0.5)

Solving for t gives:

t = -L/R * ln(0.5)

Substituting the given values L = 1H, R = 100Ω, and ln(0.5) = -0.693, we get:

t = -1/100 * -0.693 = 0.00693 s

So, the time elapsed before the current acquires half of its steady state value is approximately 0.00693 seconds.

(B) The energy stored in the magnetic field associated with the coil at instant 15 m s after the circuit is switched on.

The energy (W) stored in the magnetic field of an inductor is given by the formula:

W = 0.5 * L * I^2

We already know that the current at any time 't' is given by:

I = E/R * (1 - e^(-Rt/L))

Substituting the given values E = 6V, R = 100Ω, L = 1H, and t = 15ms = 0.015s, we get:

I = 6/100 * (1 - e^(-100*0.015))

I = 0.06 * (1 - e^-1.5)

Given that e^-1.5 = 0.25, we get:

I = 0.06 * (1 - 0.25) = 0.06 * 0.75 = 0.045 A

Substituting I = 0.045A and L = 1H into the formula for W gives:

W = 0.5 * 1 * (0.045)^2 = 0.0010125 J

So, the energy stored in the magnetic field associated with the coil at instant 15 ms after the circuit is switched on is approximately 0.0010125 Joules.

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