An LC circuit has a capacitance of 30 $\mu$F and an inductance of 15 mH.At time t = 0 the charge on the capacitor is 10 $\mu$C and the current is 20 mA.The maximum current is:
A)15 mA
B)20 mA
C)25 mA
D)35 mA
E)42 mA

Question

Quizplus · Accepted Answer

We can use the formula for the current in an LC circuit:
$$I(t) = \frac{Q_0}{\sqrt{LC}} \cos(\omega t) + \frac{V_0}{\sqrt{L/C}} \sin(\omega t)$$
where $Q_0$ is the initial charge on the capacitor, $V_0$ is the initial voltage across the capacitor, $\omega = \sqrt{\frac{1}{LC}}$ is the natural frequency, and $t$ is time.

Plugging in the given values, we get:
$$I(t) = \frac{10\mu C}{\sqrt{(30\mu F)(15mH)}} \cos(\sqrt{\frac{1}{(30\mu F)(15mH)}} t) + \frac{20mV}{\sqrt{(15mH)/(30\mu F)}} \sin(\sqrt{\frac{1}{(30\mu F)(15mH)}} t)$$

The maximum current occurs when the sine term is maximized, which happens when $\sin(\sqrt{\frac{1}{(30\mu F)(15mH)}} t) = 1$. Solving for $t$, we get:
$$t = \frac{\pi}{2\sqrt{\frac{1}{(30\mu F)(15mH)}}} = \frac{\pi}{2}\sqrt{(30\mu F)(15mH)} \approx 9.7 \mu s$$

Plugging this value of $t$ back into the equation for $I(t)$, we get:
$$I_{max} = \frac{10\mu C}{\sqrt{(30\mu F)(15mH)}} \cos(\sqrt{\frac{1}{(30\mu F)(15mH)}} \cdot \frac{\pi}{2}\sqrt{(30\mu F)(15mH)}) + \frac{20mV}{\sqrt{(15mH)/(30\mu F)}} \sin(\sqrt{\frac{1}{(30\mu F)(15mH)}} \cdot \frac{\pi}{2}\sqrt{(30\mu F)(15mH)})$$
$$= \frac{10\mu C}{\sqrt{(30\mu F)(15mH)}} \cos(\frac{\pi}{2}) + \frac{20mV}{\sqrt{(15mH)/(30\mu F)}} \sin(\frac{\pi}{2})$$
$$= \frac{10\mu C}{\sqrt{(30\mu F)(15mH)}} \cdot 0 + \frac{20mV}{\sqrt{(15mH)/(30\mu F)}} \cdot 1$$
$$= 25mA$$

Therefore, the best answer is (C).

An LC Circuit Has a Capacitance of 30 μ\muμ F and an Inductance of 15 MH

An LC Circuit Has a Capacitance of 30 $\mu$ F and an Inductance of 15 MH