Question

At , the current in the$\mathrm{t}=0\mathrm{s}$circuit in FIGURE $\mathrm{EX}30.35$is ${\mathrm{I}}_0$. At what time is the current $\frac{1}{2}{\mathrm{I}}_0$

Short answer

Time for current$\frac{1}{2}{\mathrm{I}}_0$is$173\mathrm{\mu s}$.

Step-by-step solution

Formula for current and time constant

Current decays across the inductor in a $\mathrm{LR}$circuit.

$\mathrm{I}={\mathrm{I}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}$

localid="1648833013057" $\mathrm{\tau }$is a constant in time.

Furthermore, the temporal constant is given by

$\mathrm{\tau }=\frac{\mathrm{L}}{\mathrm{R}}$

Calculation for time for current12I0

As it stands,

${\mathrm{R}}_{\mathrm{eq}}=300\mathrm{\Omega }$

$\mathrm{L}=75\times {10}^{-3}\mathrm{H}$

all values should be substituted,

$\mathrm{\tau }=\frac{\mathrm{L}}{\mathrm{R}}=\frac{75\times {10}^{-3}}{300\mathrm{\Omega }}$

$\mathrm{I}=\frac{{\mathrm{I}}_{\mathrm{o}}}{2}$Then,

$\mathrm{I}={\mathrm{I}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{\tau }}$

$\frac{{\mathrm{I}}_{\mathrm{o}}}{2}={\mathrm{I}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/250\mathrm{\mu s}}$

$\ln 0.5=\frac{-\mathrm{t}}{250\mathrm{\mu s}}$

$\mathrm{t}=173\mathrm{\mu s}$