An ideal inductor-resistor-battery circuit is switched on at $t=0$ s. At time $t$, the current is $i=i_0(1-e^{

Question

An ideal inductor-resistor-battery circuit is switched on at $t=0$ s. At time $t$, the current is $i=i_0(1-e^{-t/	au})$, where $i_0$ is the steady-state value. The time at which the current becomes $0.5i_0$ is: [Given $\ln(2)=0.693$]

Accepted Answer

The growth of current in an LR circuit is governed by the equation:
$i = i_0 (1 - e^{-t/	au})$

**1. Set the Condition:**
We need to find the time $t$ when the current $i$ reaches 50% of its steady-state value $i_0$. 
$0.5 i_0 = i_0 (1 - e^{-t/	au})$

**2. Simplify:**
$0.5 = 1 - e^{-t/	au}$
$e^{-t/	au} = 1 - 0.5 = 0.5$

**3. Solve for t:**
Taking the natural logarithm (ln) on both sides:
$-t/	au = \ln(0.5)$
$-t/	au = -\ln(2)$
$t = 	au \ln(2)$

**4. Calculation:**
Given $\ln(2) = 0.693$, we have:
$t = 0.693 	au$

*(Note: The provided question text is missing the specific value of the time constant $	au$. However, for the answer to be 6.93 ms, the time constant $	au$ must be 10 ms. Assuming $	au = 10 	ext{ ms}$, then $t = 0.693 	imes 10 	ext{ ms} = 6.93 	ext{ ms}$.)*

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