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$t=40s$ is close to take $e^2=7.389$.

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

Answer 1

The correct option is A

$1.06$

Step 1. Given data

EMF$=20V$

Time, $t=0s$

$e^2=7.389$

Resistor, $R=5\Omega$

Step 2. Finding the ratio of the currents

$i=i_0\left(1-e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$\tau $}\right.}\right)$

Where, $\tau$ is torque

$t$ is time

$i_0$ is initial current

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

Where, $\tau$ is torque

$L$ is inductance

$R$ is resistance

$\begin{array}{rcl}i& =& i_0\left(1-e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$\tau $}\right.}\right)\\ L& =& 0.01\\ i_0& =& \frac{emf}{R}\\ i& =& \frac{20}{5}\left(1-e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \raisebox{1ex}{$0.01$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$}\right.}\right)\\ i& =& 4\left(1-e^{-500t}\right)\end{array}$

So,

$i_{\infty }=4$

$$\begin{gathered} i_{40}=4(1-e^{-500\times 40}) \\ {i}_{40}=4\left(1-\frac{1}{{\left(e^2\right)}^{100}}\right) \\ {i}_{40}=4\left(1-\frac{1}{7.38910000}\right) \\ \frac{i_{\infty }}{i_{40}}\approx 1 \end{gathered}$$ [Slightly greater than 1]

Hence, the correct option is $\left(A\right)$.