A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz.At time t = 0 the capacitor is fully charged.At a given instant later the charge on the capacitor is measured to be 3.0 μC and the current in the circuit is equal to 75 μA.What is the maximum charge of the capacitor?
A) 6.0 µC
B) 10 µC
C) 8.0 µC
D) 2.0 µC
E) 4.0 µC

Question

Quizplus · Accepted Answer

At any time in an LC circuit, the charge on the capacitor can be found using the equation:

$Q = Q_{max}\cos(\omega t)$

where $Q_{max}$ is the maximum charge on the capacitor and $\omega$ is the angular frequency of the circuit.

The angular frequency can be found using the formula:

$\omega = 2\pi f$

where $f$ is the frequency of oscillation given in the problem.

At the instant in question, the charge on the capacitor is measured to be 3.0 µC and the current is 75 µA. We know that the current in an LC circuit is given by:

$I = \frac{Q}{L}\omega\sin(\omega t)$

where $L$ is the inductance of the circuit. We can solve for $L$ using this equation:

$L = \frac{Q}{I\omega\sin(\omega t)}$

Using the given values, we get:

$L = \frac{3.0\mu C}{75\mu A	imes 2\pi	imes 1.6 Hz	imes\sin(2\pi	imes 1.6 Hz	imes t)}$

Simplifying this gives:

$L = \frac{2}{5\pi	imes\sin(3.2\pi	imes t)}$

We can use this expression for $L$ to find $Q_{max}$. At the point of maximum charge on the capacitor, the current is zero, so we can set $I = 0$ in the equation above:

$L_{max} = \frac{Q_{max}}{\omega\sin(\omega t_{max})}$

Solving for $Q_{max}$, we get:

$Q_{max} = L_{max}\omega\sin(\omega t_{max})$

Now we need to find the time $t_{max}$ when the capacitor has its maximum charge. We know that at time $t = 0$, the capacitor is fully charged, so the first time when it is fully charged again (i.e. when $Q = Q_{max}$) must be some integer multiple of half the period $T = 1/f = 0.625$ s later.

$Q = Q_{max}\cos(\omega t_{max}) = Q_{max}$

$\cos(\omega t_{max}) = 1$

$\omega t_{max} = 2n\pi$ where $n$ is an integer

$t_{max} = \frac{2n\pi}{\omega}$

We want the first non-zero value of $t_{max}$, so we take $n = 1$:

$t_{max} = \frac{2\pi}{\omega} = \frac{1}{f} = 0.625$ s

Now we can finally calculate $Q_{max}$:

$Q_{max} = L_{max}\omega\sin(\omega t_{max}) = \frac{2}{5\pi}	imes 2\pi f	imes\sin(2\pi f	imes 0.625) \approx 8.0\mu C$

Therefore, the maximum charge on the capacitor is approximately 8.0 µC, so the answer is (C).

A Charged Capacitor Is Connected to an Ideal Inductor to Form