Question

The switch in FIGURE EX30.33 has been in position 1 for a long time. It is changed to position 2 at t = 0 s. a. What is the maximum current through the inductor? b. What is the first time at which the current is maximum?

Short answer

The maximum current through inductor is $0.24\times {10}^{-4}$A and at $4.96\times {10}^{-4}$s current was maximum for first time.

Step-by-step solution

Given information

We have given that, the switch in FIGURE EX30.33 has been in position 1 for a long time. It is changed to position 2 at t = 0 s.

Part (a) Step 1.

The formula to calculate the maximum current :

$\frac{1}{2}C{V}^2=\frac{1}{2}L{I}^2$

$I=\sqrt{\frac{C{V}^2}{L}}$

Here, I = Current

C = Capacitance

V = Potential difference

L = Inductance

Step 2. 

Substituting the values in the formula :

$I=\sqrt{\frac{(2\times {10}^{-6}F)\times {(12V)}^2}{50\times {10}^{-3}H}}$

$I=0.24\times {10}^{-4}$A

Part (b) Step 1. 

We know that $T=\frac{2\pi }{\omega }$,and ${\omega }^2=\frac{1}{LC}$

So, the first time at which the current is maximum :

$t=\frac{T}{4}$

$t=\frac{2\times \pi \times \sqrt{LC}}{4}$

Here, t = Time at which current was maximum for the first time

L = Inductance

C = Capacitance

Step 2. 

Substituting the values in the formula of t :

$t=\frac{2\times 3.14\times \sqrt{(50\times {10}^{-3}H)\times (2\times {10}^{-6}F)}}{4}$

$t=4.96\times {10}^{-4}$s