Electromagnetic Induction and Alternating Current 4 Question 20
####20. An inductor of inductance $2.0 mH$ is connected across a charged capacitor of capacitance $5.0 \mu F$ and the resulting $L-C$ circuit is set oscillating at its natural frequency. Let $Q$ denote the instantaneous charge on the capacitor and $I$, the current in the circuit. It is found that the maximum value of $Q$ is $200 \mu C$.
$(1998,8 M)$
(a) When $Q=100 \mu C$, what is the value of $|d I / d t|$ ?
(b) When $Q=200 \mu C$, what is the value of $I$ ?
(c) Find the maximum value of $I$.
(d) When $I$ is equal to one-half its maximum value, what is the value of $|Q|$ ?
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Answer:
Correct Answer: 20. $\text { (a) } 10^4 A / s \text { (b) zero (c) } 2.0 A \text { (d) } 1.732 \times 10^{-4} C$
Solution:
- This is a problem of $L-C$ oscillations. Charge stored in the capacitor oscillates simple harmonically as
$$ Q=Q _0 \sin (\omega t \pm \varphi) $$
Here, $Q _0=$ maximum value of $Q=200 \mu C=2 \times 10^{-4} C$
$$ \omega=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{\left(2 \times 10^{-3}\right)\left(5.0 \times 10^{-6}\right)}}=10^{4} s^{-1} $$
Let at
$$ t=0, Q=Q _0 \text {, then } $$
$$ \begin{aligned} & Q(t)=Q _0 \cos \omega t \\ & I(t)=\frac{d Q}{d t}=-Q _0 \omega \sin \omega t \text { and } \end{aligned} $$
$\frac{d I(t)}{d t}=-Q _0 \omega^{2} \cos (\omega t)$
(a) $Q=100 \mu C$ or $\frac{Q _0}{2}$ at $\cos \omega t=\frac{1}{2}$ or $\omega t=\frac{\pi}{3}$
At $\cos (\omega t)=\frac{1}{2}$, from Eq. (iii) :
$$ \begin{aligned} & \frac{d I}{d t}=\left(2.0 \times 10^{-4} C\right)\left(10^{4} s^{-1}\right)^{2} \frac{1}{2} \\ & \frac{d I}{d t}=10^{4} A / s \end{aligned} $$
(b) $Q=200 \mu C$ or $Q _0$ when $\cos (\omega t)=1$ i.e. $\omega t=0,2 \pi \ldots$
At this time $I(t)=-Q _0 \omega \sin \omega t$
$$ \text { or } \quad I(t)=0 \quad\left(\sin 0^{\circ}=\sin 2 \pi=0\right) $$
(c) $I(t)=-Q _0 \omega \sin \omega t$
$\therefore$ Maximum value of $I$ is $Q _0 \omega$.
$$ \begin{aligned} & I _{\max }=Q _0 \omega=\left(2.0 \times 10^{-4}\right)\left(10^{4}\right) \\ & I _{\max }=2.0 A \end{aligned} $$
(d) From energy conservation,
$$ \begin{aligned} \frac{1}{2} L I _{\max }^{2} & =\frac{1}{2} L I^{2}+\frac{1}{2} \frac{Q^{2}}{C} \\ \text { or } \quad Q & =\sqrt{L C\left(I _{\max }^{2}-I^{2}\right)} \\ I & =\frac{I _{\max }}{2}=1.0 A \\ \therefore \quad Q & =\sqrt{\left(2.0 \times 10^{-3)}\left(5.0 \times 10^{-6}\right)\left(2^{2}-1^{2}\right)\right.} \\ Q & =\sqrt{3} \times 10^{-4} C \text { or } Q=1.732 \times 10^{-4} C \end{aligned} $$
