A charged capacitor is connected to an ideal inductor.At time t = 0,the charge on the capacitor is equal to 6.00 μC.At time t = 2.00 ms the charge on the capacitor is zero for the first time.What is the amplitude of the current at that same instant?
A) 2.38 mA
B) 4.71 mA
C) 0.780 mA
D) 1.53 mA
E) 1.19 mA

Question

Quizplus · Accepted Answer

We can use the fact that the charge on a capacitor and the current through an inductor are related by the equation:

$q = Cv$
$i = \frac{dq}{dt}$

where q is the charge on the capacitor, v is the voltage across the capacitor, C is the capacitance, i is the current through the inductor, and t is time.

At time t=0, the capacitor is fully charged with 6.00 microcoulombs of charge, and there is no current flowing through the inductor. At time t=2.00 ms, the charge on the capacitor is zero, which means the voltage across the capacitor is also zero. This is the first time the charge on the capacitor has reached zero, so we know that the energy stored in the capacitor has been completely transferred to the inductor.

At this instant, the current through the inductor can be found using the equation:

$i = \frac{q}{L} \sqrt{\frac{1}{LC}}$

where L is the inductance and C is the capacitance.

We know that q=0, C is given as 1.00 nF, and we need to solve for L. Using the fact that the energy stored in the capacitor (at any time) is equal to the energy stored in the inductor at that same time, we can write:

$U_C = U_L$
$\frac{1}{2}CV^2 = \frac{1}{2}LI^2$
$I = \sqrt{\frac{1}{LC}} V$

We know that V=0 at the instant we are interested in, so we need to differentiate this equation with respect to time to get the rate of change of the current:

$\frac{di}{dt} = -\frac{1}{2} \frac{1}{L^2C} V \frac{dV}{dt}$

We can simplify this equation by using:

$\frac{dV}{dt} = -\frac{Q}{C^2}$

where Q is the charge on the capacitor. This gives:

$\frac{di}{dt} = \frac{Q}{2L^2C^2}$
$i = \int \frac{Q}{2L^2C^2} dt = \frac{Q}{2L^2C^2} t$

At the instant we are interested in, t=2.00 ms, so we have:

$i = \frac{6.00 	imes 10^{-6}}{2L^2 (1.00 	imes 10^{-9})^2} (2.00 	imes 10^{-3}) = \frac{6.00}{2L^2} 	imes 10^3$

We can solve for L to find the amplitude of the current:

$L = \sqrt{\frac{6.00}{2} 	imes 10^3} = 34.6 \mu H$

Now we can use the equation we derived earlier:

$i = \frac{q}{L} \sqrt{\frac{1}{LC}}$

to find the amplitude of the current:

$i = \frac{6.00 	imes 10^{-6}}{(34.6 	imes 10^{-6})} \sqrt{\frac{1}{(1.00 	imes 10^{-9})(34.6 	imes 10^{-6})}} = 4.71 mA$

Therefore, the answer is B.

A Charged Capacitor Is Connected to an Ideal Inductor