Question

A \(4.00-\mathrm{mF}\) capacitor is connected in series with a \(7.00-\mathrm{mH}\) inductor. The peak current in the wires between the capacitor and the inductor is \(3.00 \mathrm{~A}\). a) What is the total electric energy in this circuit? b) Write an expression for the charge on the capacitor as a function of time, assuming the capacitor is fully charged at \(t=0 \mathrm{~s}\).

Short answer

a) The total electric energy in the circuit is given by: $$U_{total} = \frac{1}{2}(4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A))^2$$ b) The expression for the charge on the capacitor as a function of time is: $$Q(t) = (4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A)) \cos\left(\frac{1}{\sqrt{(4.00\times 10^{-3} F)(7.00\times 10^{-3} H)}}t\right)$$

Step-by-step solution

Find the initial energy stored in the capacitor and the inductor

At \(t = 0s\), the capacitor is fully charged, and the inductor has no current passing through it. The energy stored in the inductor, \(U_L\), can be found using the formula \(U_L = \frac{1}{2}LI^2\), where \(L\) is the inductance and \(I\) is the current through the inductor. Since at \(t=0\), \(I=0\), we have: $$U_L = \frac{1}{2}(7.00\times 10^{-3} H)(0 A)^2 = 0 J$$ The energy stored in the capacitor, \(U_C\), can be found using the formula \(U_C = \frac{1}{2}CV^2\), where \(C\) is the capacitance and \(V\) is the voltage across the capacitor. At \(t=0\), the peak current (\(I_0\)) is 3.00 A, which implies the maximum voltage (\(V_0\)) across the inductor is \(V_0 = L\cdot I_0\), so: $$V_0 = (7.00\times10^{-3} H)(3.00 A)$$ Since the capacitor and inductor are in series, the voltage across the capacitor (\(V_C\)) is also \(V_0\): $$V_C = V_0 = (7.00\times10^{-3} H)(3.00 A)$$ Now we can find the energy stored in the capacitor: $$U_C = \frac{1}{2}(4.00\times 10^{-3} F)V_C^2$$ $$U_C = \frac{1}{2}(4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A))^2$$

Calculate the total electric energy in the circuit

The total electric energy in the circuit is the sum of the energy stored in the capacitor and the energy stored in the inductor, so: $$U_{total} = U_C + U_L = U_C = \frac{1}{2}(4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A))^2$$

Write down the expression for the charge on the capacitor as a function of time

We can use the formula \(Q(t) = Q_0 \cos(\omega t)\), where \(Q_0\) is the initial charge on the capacitor and \(\omega\) is the angular frequency of oscillation. The angular frequency of oscillation in the LC circuit is given by: $$\omega = \frac{1}{\sqrt{LC}}$$ We can find the initial charge on the capacitor (\(Q_0\)) using the initial voltage across the capacitor: $$Q_0 = CV_C = (4.00\times 10^{-3} F)(V_0) = (4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A))$$ So, the expression for the charge on the capacitor as a function of time is: $$Q(t) = (4.00\times 10^{-3} F)((7.00\times 10^{-3} H)(3.00 A)) \cos\left(\frac{1}{\sqrt{(4.00\times 10^{-3} F)(7.00\times 10^{-3} H)}}t\right)$$