Question

What is the inductance of a series RL circuit in which R = 1.0 kΩ if the current increases to one-third of its final value in 30 μs? A) 74 mH B) 99 mH C) 49 mH D) 62 mH E) None of the above

Answer 1

The inductance of a series RL circuit can be calculated using the formula: L = (V * t)/(ΔI). In this case, with a resistance of 1.0 kΩ and a current increase of one-third of its final value in 30 μs, the inductance is 49 mH.

Step-by-step explanation:

The inductance of a series RL circuit can be calculated using the formula:
L = (V * t)/(ΔI)
Where L is the inductance in henries, V is the voltage across the inductor in volts, t is the time in seconds, and ΔI is the change in current in amperes.

In this case, the current increases to one-third of its final value, which means ΔI is 2/3 of the final current. The time is given as 30 μs and the resistance is 1.0 kΩ, which is equivalent to 1000 ohms.

Using the formula, we can calculate the inductance:
L = (V * t)/(ΔI) = (1000 * (30 * 10^-6))/(2/3) = 49 mH

Therefore, the correct answer is C) 49 mH.

Answer 2

The inductance of the series RL circuit is approximately 74 mH.

The inductance of a series RL circuit can be calculated using the formula

`L = (R * T)/(ln((1-I)/(1-I_final))).`

Given that

R = 1.0 kΩ,

T = 30 μs,

I = 0, and
`I_final` =1/3

we can substitute these values into the formula.

Plugging in these values, we get

`L = (1000 * 30 * 10^-6)/(ln((1-0)/(1-1/3)))`

`\[ L = (1000 * 30 * 10^(-6))/(\ln\left((1-0)/(1-(1)/(3))\right)) \]`

First, simplify the expression within the natural logarithm:

`\[ (1-0)/(1-(1)/(3)) = (1)/((2)/(3)) = (3)/(2) \]`

Now, substitute this result back into the original expression:

`\[ L = (1000 * 30 * 10^(-6))/(\ln\left((3)/(2)\right)) \]`

To evaluate the natural logarithm term, you can use the fact that
`\( \ln((a)/(b)) = \ln(a) - \ln(b) \)`

`\[ L = (1000 * 30 * 10^(-6))/(\ln(3) - \ln(2)) \]`

The correct answer is A) 74 mH.