A certain circuit consists of an inductor of

50 mH in series with a resistor of 130 Ω. At
one moment, the current in the circuit is 12 A,
and decreasing.
How long will it take for the current to fall
to 4.8 A?
Answer in units of s.

I assume a short circuit so total voltage is 0

V = L di/dt + iR = 0

let i = I e^kt
then when t =0, i = I
di/dt = I k e^kt = k i

V = L I k e^kt + I e^kt R = 0
L k + R = 0
k = -R/L but I bet you knew that

so
k = -130/50*10^-3 = -2.6*10^3
I = 12
so
i = 12 e^-2.6*10^3 t
4.8/12 = .4
ln .4 = -.916 = -2.6*10^3 t
solve for t (not long :)

To determine the time it takes for the current to fall to 4.8 A in the given circuit, we can use the formula for the time constant (τ) of an RL circuit:

τ = L / R

where:
L = inductance (50 mH = 0.05 H)
R = resistance (130 Ω)

Plugging in the values, we can calculate the time constant:

τ = 0.05 H / 130 Ω

Simplifying the expression:

τ = 0.000384615 s

The time constant represents the time it takes for the current to decrease to approximately 36.8% of its initial value. By using this information, we can determine the time it takes for the current to fall to 4.8 A.

To find the time (t) in seconds, we can use the formula:

t = τ * ln(Iinitial / Ifinal)

where:
Iinitial = initial current (12 A)
Ifinal = final current (4.8 A)
ln = natural logarithm

Plugging in the values:

t = 0.000384615 s * ln(12 A / 4.8 A)

Calculating ln(12 A / 4.8 A):

t = 0.000384615 s * ln(2.5)

Using a calculator to evaluate ln(2.5), we can find:

t ≈ 0.000384615 s * 0.916

Calculating the product:

t ≈ 0.0003527 s

Therefore, it will take approximately 0.0003527 seconds (or 3.527 milliseconds) for the current to fall to 4.8 A in the given circuit.

To find out how long it will take for the current to fall to 4.8 A in the given circuit, we first need to understand the behavior of an RL circuit.

In an RL circuit, the current through the circuit changes over time due to the presence of inductance. The rate of change of current is determined by the time constant, denoted as τ (tau).

The time constant (τ) of an RL circuit can be calculated using the formula:

τ = L / R

Where:
τ = Time constant (in seconds)
L = Inductance (in henries)
R = Resistance (in ohms)

Given:
L = 50 mH = 50 × 10^(-3) H
R = 130 Ω

We can now substitute these values into the formula to find the time constant:

τ = (50 × 10^(-3)) / 130
= 3.846 × 10^(-4) s

The time constant represents the time required for the current to decrease to approximately 36.8% of its initial value. We can now use this information to find the time it takes for the current to drop from 12 A to 4.8 A.

Let's consider Δt as the time taken for the current to decrease from 12 A to 4.8 A.

Using the exponential decay equation for an RL circuit:
I(t) = I0 × e^(-t/τ)

Where:
I(t) = Current at time t
I0 = Initial current
t = Time (in seconds)
τ = Time constant

We know that I(t) = 4.8 A and I0 = 12 A. Plugging in these values in the equation and solving for t:

4.8 = 12 × e^(-t/τ)

Dividing both sides by 12:

0.4 = e^(-t/τ)

Taking the natural logarithm (ln) of both sides:

ln(0.4) = -t/τ

Rearranging the equation to solve for t:

t = -τ × ln(0.4)

Substituting the value of τ:
t = -3.846 × 10^(-4) s × ln(0.4)
t ≈ 3.846 × 10^(-4) s × 0.916

Calculating the value using a calculator:
t ≈ 3.527 × 10^(-4) s

Therefore, it will take approximately 3.527 × 10^(-4) seconds (or 0.3527 ms) for the current to fall from 12 A to 4.8 A in this circuit.