Circuits with Capacitors and Inductors
Help Questions
AP Physics C: Electricity and Magnetism › Circuits with Capacitors and Inductors
An ideal LC circuit has a total electromagnetic energy of $$50$$ $$\mu$$J. At a certain instant, the energy stored in the electric field of the capacitor is $$30$$ $$\mu$$J. What is the energy stored in the magnetic field of the inductor at that same instant?
$$80$$ $$\mu$$J
$$50$$ $$\mu$$J
$$30$$ $$\mu$$J
$$20$$ $$\mu$$J
Explanation
In an ideal LC circuit, total energy is conserved. The total energy $$U_{total}$$ is the sum of the energy in the capacitor $$U_C$$ and the energy in the inductor $$U_L$$. Therefore, $$U_L = U_{total} - U_C = 50 \mu J - 30 \mu J = 20 \mu J$$.
In an oscillating ideal LC circuit, the total energy is constant. At the instant the charge on the capacitor is at its maximum positive value, which of the following statements is true?
The energy stored in the inductor's magnetic field is at its maximum value.
The energy is equally shared between the capacitor and the inductor.
The current flowing through the inductor is at its maximum value.
The current flowing through the inductor is zero.
Explanation
When the charge on the capacitor is maximum, all the energy of the circuit is stored in the capacitor's electric field. Therefore, the energy stored in the inductor is zero. Since inductor energy is given by $$\frac{1}{2}LI^2$$, the current $$I$$ must be zero at this instant.
In an ideal LC circuit, the charge on the capacitor $$q(t)$$ and the current in the inductor $$I(t)$$ oscillate sinusoidally. What is the phase difference between the charge oscillations and the current oscillations?
They are out of phase by $$\pi/2$$ radians ($$90^\circ$$).
They are out of phase by $$\pi/4$$ radians ($$45^\circ$$).
They are out of phase by $$\pi$$ radians ($$180^\circ$$).
They are in phase ($$0$$ radians).
Explanation
If the charge is given by $$q(t) = Q_{max} \cos(\omega t)$$, the current is $$I(t) = dq/dt = -\omega Q_{max} \sin(\omega t)$$. The sine and cosine functions are out of phase by $$\pi/2$$ radians. The current leads the charge by $$\pi/2$$ radians.
An LC circuit is oscillating. At the instant when the capacitor stores its maximum energy, the plates of the capacitor are pulled apart, doubling their separation distance. How does the maximum current $$I_{max}$$ in the subsequent oscillations compare to its initial value?
It increases because the total energy of the circuit increases.
It remains the same because the charge on the capacitor is conserved.
It decreases because the capacitance decreases, reducing the stored charge.
It decreases because the total energy of the circuit decreases.
Explanation
At the instant of maximum energy in the capacitor, the current is zero, so the capacitor is electrically isolated. Pulling the plates apart does positive work on the system, which increases the stored potential energy $$U_C = Q^2/(2C)$$ because C decreases while Q remains constant. This increased total energy is then conserved. Since the new maximum inductor energy $$\frac{1}{2}LI_{new,max}^2$$ must equal this new, larger total energy, the new maximum current must be greater than the initial maximum current.
In an ideal LC circuit, a capacitor with capacitance $$C$$ is initially charged to a maximum charge $$Q_{max}$$. The circuit then oscillates. By conservation of energy, what is the maximum energy $$U_{L,max}$$ stored in the inductor during one cycle?
$$U_{L,max} = \frac{Q_{max}^2}{2L}$$
$$U_{L,max} = 0$$
$$U_{L,max} = \frac{1}{2} L Q_{max}^2$$
$$U_{L,max} = \frac{Q_{max}^2}{2C}$$
Explanation
The total energy in an ideal LC circuit is conserved. It oscillates between the capacitor's electric field and the inductor's magnetic field. The maximum energy stored in the inductor must equal the maximum energy initially stored in the capacitor, which is $$U_{C,max} = \frac{Q_{max}^2}{2C}$$.
An ideal LC circuit consists of a capacitor with capacitance $$C$$ and an inductor with inductance $$L$$. What is the angular frequency $$\omega$$ of the electromagnetic oscillations in the circuit?
$$\omega = \frac{1}{\sqrt{LC}}$$
$$\omega = \sqrt{\frac{L}{C}}$$
$$\omega = \sqrt{LC}$$
$$\omega = \sqrt{\frac{C}{L}}$$
Explanation
The differential equation for an ideal LC circuit is $$L\frac{d^2q}{dt^2} + \frac{1}{C}q = 0$$, which is the equation for simple harmonic motion, $$\frac{d^2q}{dt^2} + \omega^2 q = 0$$. Comparing these gives $$\omega^2 = \frac{1}{LC}$$, so the angular frequency is $$\omega = \frac{1}{\sqrt{LC}}$$.
An ideal LC circuit contains an inductor of inductance $$L$$ and a capacitor of capacitance $$C$$. What is the period $$T$$ of the oscillations in this circuit?
$$T = \frac{\sqrt{LC}}{2\pi}$$
$$T = 2\pi\sqrt{LC}$$
$$T = 2\pi\sqrt{\frac{L}{C}}$$
$$T = \frac{1}{2\pi\sqrt{LC}}$$
Explanation
The angular frequency of an LC circuit is $$\omega = \frac{1}{\sqrt{LC}}$$. The period $$T$$ is related to the angular frequency by $$T = \frac{2\pi}{\omega}$$. Substituting the expression for $$\omega$$ gives $$T = 2\pi\sqrt{LC}$$.
An LC circuit has an oscillation period of $$T$$. If the inductance $$L$$ is quadrupled while the capacitance $$C$$ remains the same, what is the new period of oscillation?
$$4T$$
$$T/2$$
$$2T$$
$$T$$
Explanation
The period of an LC circuit is given by $$T = 2\pi\sqrt{LC}$$. If the inductance becomes $$4L$$, the new period $$T_{new}$$ will be $$T_{new} = 2\pi\sqrt{(4L)C} = 2(2\pi\sqrt{LC}) = 2T$$.
An ideal LC circuit has a capacitor with capacitance $$C$$ initially charged to a potential difference $$V_{max}$$. The maximum current is $$I_{max}$$. If the capacitance is changed to $$4C$$ and the capacitor is again charged to the same initial potential difference $$V_{max}$$, what is the new maximum current?
$$4I_{max}$$
$$2I_{max}$$
$$I_{max}$$
$$I_{max}/2$$
Explanation
By energy conservation, the maximum energy in the capacitor equals the maximum energy in the inductor. Initially, $$\frac{1}{2}CV_{max}^2 = \frac{1}{2}LI_{max}^2$$. With the new capacitance, $$\frac{1}{2}(4C)V_{max}^2 = \frac{1}{2}LI_{new,max}^2$$. This means $$4(\frac{1}{2}CV_{max}^2) = \frac{1}{2}LI_{new,max}^2$$, so $$4(\frac{1}{2}LI_{max}^2) = \frac{1}{2}LI_{new,max}^2$$. This simplifies to $$4I_{max}^2 = I_{new,max}^2$$, so $$I_{new,max} = 2I_{max}$$.
An ideal LC circuit consists of an inductor with inductance $$L$$ and a capacitor with capacitance $$C$$. Applying Kirchhoff's loop rule to this circuit results in which of the following differential equations, where $$q$$ is the charge on the capacitor and $$I$$ is the current?
$$I + C\frac{dq}{dt} = 0$$
$$\frac{dI}{dt} + LCq = 0$$
$$L\frac{dI}{dt} - \frac{q}{C} = 0$$
$$L\frac{dI}{dt} + \frac{q}{C} = 0$$
Explanation
According to Kirchhoff's loop rule, the sum of potential differences around a closed loop is zero. The potential difference across the inductor is $$V_L = L\frac{dI}{dt}$$ and across the capacitor is $$V_C = \frac{q}{C}$$. Therefore, $$V_L + V_C = 0$$, which gives $$L\frac{dI}{dt} + \frac{q}{C} = 0$$.