Question

A 35.0-V battery with negligible internal resistance, a 50.0-Ω resistor, and a 1.25-mH inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed.How long after closing the switch will the current through the inductor reach one-half of its maximum value?

Answer 1

Answer:$t=1.732\times 10^{-5}\ s$

Explanation:

Given

Voltage $V=35\ V$

Resistance $R=50\ Omega$

Inductance $L=1.25\ mH$

Current in the L-R circuit increases as

$i=\frac{V}{R}(1-e^{-\frac{t}{\tau }})$

Where $\tau$=time constant $(\frac{L}{R})$

$\frac{V}{R}=i$(Maximum current)

Time at which current will be half of maximum i.e.

$(1-e^{-\frac{t}{\tau }})=\frac{1}{2}$

$e^{-\frac{t}{\tau }}=\frac{1}{2}$

$\frac{t}{\tau }=\ln (2)$

$t=\frac{L}{R}\times \ln (2)$

$t=\frac{1.25\times 10^{-3}}{50}\times \ln (2)$

$t=1.732\times 10^{-5}\ s$

Answer 2

After closing the switch the current through the inductor reach one-half of its maximum value in "1.732 × 10⁻⁵ s".

L-R Circuit

According to the question,

Voltage, V = 35 V

Resistance, R = 50.0 Ω

Inductance, L = 1.25 mH

Time constant, $\tau$ = $\frac{L}{R}$

Max. current, i = $\frac{V}{R}$

We know the relation,

→ i = $\frac{V}{R} (1 - e^{-\frac{t}{\tau} })$

or,

 (1 - $e^{-\frac{t}{r} }$) = $\frac{1}{2}$

       $e^{-\frac{t}{r} }$ = $\frac{1}{2}$

           $\frac{t}{\tau}$ = ln(2)

or, t = $\frac{L}{R}$ × ln(2)

By substituting the values,

       = $\frac{1.25\times 10^{-3}}{50}$ × ln(2)  

       = 1.732 × 10⁻⁵ s

Thus the above answer is right.

             

Find out more information about resistance here:

https://brainly.com/question/15595203