Question

An emf of $20V$ is applied at time $t=0$ to a circuit containing in series $10mH$ inductor and $5\Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40$ s is close to : (Take $e^2=7.389$)

• A. $1.06$
• B. $1.15$
• C. $0.84$
• D. $1.46$

Answer 1

Answer: (A)

$\text{ We know, }i=i_0\left(1-e^{-t/\tau }\right)$
$\tau =\frac{L}{R}=\frac{0.01}{5}=0.002$
$i=i_0\left(1-e^{\frac{-t}{0.002}}\right)=i_0\left(1-e^{-500t}\right)$
$\text{ at }t=40s$
$i_{40}=i_0\left(1-e^{-500\times 40}\right)$
$=\frac{20}{5}\left(1-e^{-20000}\right)$
$i_{40}=4\left(1-e^{-20000}\right)\left[\because i_0=\frac{20v}{5\omega }\right]$
$\text{ at }t=\infty$
$i_{\infty }=i_0\left(1-e^{-t/\tau }\right)=i_0\left(1-e^{-\infty }\right)$
$i_{\infty }=\frac{20}{5}(1-0)=4$
$\Rightarrow \frac{i_{\infty }}{i_{40}}=\frac{4}{4\left(1-e^{-20000}\right)}=\frac{1}{\left(1-e^{-20000}\right)}$
$e^{-20000}\simeq 0\&1>\left(1-e^{-20000}\right)$
$\text{ So the most probable answer is }1.06$