Irodov 3.317: Halfway After 0.69 Time Constants – Amit & Deepali Sinha Foundation

Problem Irodov 3.317

A coil with inductance $L = 300$ mH and resistance $R = 140$ m $\Omega$ is connected to a source of constant voltage. How long after the connection does the current in the coil reach the fraction $\eta = 50\%$ of its final, steady value?

The constant voltage source V and the coil, drawn as its inductance L in series with its resistance R, before their terminals are joined. — The constant voltage source $V$ and the coil, drawn as its inductance $L$ in series with its resistance $R$ , before their terminals are joined.

Given

$L = 300$ mH
$R = 140$ m $\Omega$
$\eta = 0.50$
coil current zero at the connection (assumed)
source resistance negligible (assumed)

Find

the time

t_\eta

at which

I = \eta I_0

Solution

Step 1The equation of the current

Call the source voltage $V$ and take $t = 0$ at the connection, when the current is zero (assumed). Going round the loop, the source voltage is shared between the resistance and the self-induced emf of the coil, which opposes the growth of the current:

V = RI + L\,\frac{dI}{dt} \tag{1}

Once the current stops changing, (1) leaves only the resistance, so the steady current is

I_0 = \frac{V}{R} \tag{2}

Step 2Solving for the rise

Writing (1) in terms of the shortfall $I_0 - I$ with (2) separates the variables:

\frac{dI}{I_0 - I} = \frac{R}{L}\,dt

Integrating from the start, $I = 0$ at $t = 0$ , to a later instant:

\int_0^{I}\frac{dI'}{I_0 - I'} = \frac{R}{L}\int_0^{t}dt' \quad\Longrightarrow\quad -\ln\frac{I_0 - I}{I_0} = \frac{R}{L}\,t

I = I_0\left(1 - e^{-Rt/L}\right)

(3)

Step 3The instant the fraction is reached

The coil current rising toward its steady value I_0 with time constant L/R; the dotted level η I_0 is reached at the instant t_η marked on the time axis. — The coil current rising toward its steady value $I_0$ with time constant $L/R$ ; the dotted level $\eta I_0$ is reached at the instant $t_\eta$ marked on the time axis.

Setting $I = \eta I_0$ at the instant $t = t_\eta$ in (3) and solving for that instant:

1 - \eta = e^{-Rt_\eta/L} \quad\Longrightarrow\quad \frac{R}{L}\,t_\eta = -\ln(1 - \eta)

Numerically, with $L/R = 0.300/0.140$ s and $\eta = 0.50$ :

t_\eta = \frac{0.300\ \text{H}}{0.140\ \Omega}\,\ln 2 = 2.14\ \text{s}\times 0.693

t_\eta = -\frac{L}{R}\ln(1 - \eta) \approx 1.5\ \text{s}

(4)

Check

Units: H/ $\Omega$ is V s/A divided by V/A, a time. Limits: $\eta \to 0$ gives $t_\eta \approx \eta L/R$ , the initial slope $dI/dt = V/L$ of (1) carrying the current up to $\eta I_0 = \eta V/R$ linearly; $\eta \to 1$ gives $t_\eta \to \infty$ , since the steady value is only approached.

Remark

Half the steady current takes $\ln 2 \approx 0.69$ time constants whatever the coil, the half-life of any exponential approach. The last stretch is the slow one: reaching $99\%$ takes $\ln 100 \approx 4.6$ time constants, almost seven times as long as reaching half. Here $L/R$ is about 2 s because the resistance is only $0.14\ \Omega$ ; a coil of heavy wire is sluggish even with a modest inductance.