Question

The switch was in position (1) for long time and was moved to position (2) at $t=0$ is
$\frac{d^{2}i}{d{t}^2}$ at $t=0^{+}$

• A. $1000A/se{c}^2$
• B. $1000A/se{c}^2$
• C. $11500A/se{c}^2$
• D. $-11500A/se{c}^2$

Answer 1

The correct option is C $11500A/se{c}^2$

Switch is in position (1) in steady state


$I=\frac{5V}{2.5k\Omega }=2mA$

Now switch is moved to position (2)


Applying KVL in the circuit, we get
$iR+L\frac{di}{dt}+\frac{1}{C}\int idt=0$
At, $t=0^{+},i\left(0^{+}\right)=I=2mA$
$\frac{1}{C}\int idt=0$
$\therefore IR+L\frac{di}{dt}=0$____(i)
$\therefore \frac{di}{dt}=-\frac{IR}{L}=\frac{-2mA\times 2.5k\Omega }{1H}=-5A/sec$
Differentiating equation (i)
$R\frac{di}{dt}+L\frac{d^{2}i}{d{t}^2}+\frac{1}{C}i=0$
At, $t=0^{+}$
$2.5\times {10}^3\times (-5)+1\times \frac{d^{2}i}{d{t}^2}+\frac{2\times {10}^{-3}}{2\times {10}^{-6}}=0$
$-12.5\times {10}^3+\frac{d^{2}i}{d{t}^2}+1\times {10}^3=0$
$\frac{d^{2}i}{d{t}^2}=11.5\times {10}^{3}A/se{c}^2$
$\therefore {\begin{array}{c}\frac{d^{2}i}{d{t}^2}\end{array}|}_{t=0^{+}}=11500A/se{c}^2$