Question

Please show steps and answer.

A circuit is constructed with two capacitors and an inductor as shown. The values for the capacitors are: C1 = 324 muF and C2 = 364 muF. The inductance is L = 212 mH. At time t =0, the current through the inductor has its maximum value IL(0) = 135 mA and it has the direction shown. 1) What is wo, the resonant frequency of this circuit? radians/s 2) What is Q1(t1), the charge on the capacitor C1 at time t = t1 = 22.1 ms? The sign of Q1 is defined to be the sameas the sign of the potential difference Vab = Va - Vb at time t = t1. C 3) What is VbC(t1)= Vb - VC, the voltage across the inductor at time t1 = 22.1 ms? Note that this voltage is a signed number. V 4) What is Q1,max, the magnitude of the maximum charge on capacitor C1?

Answer 1

We transformed the original ciruit combining the capacitors to obtain a LC circuit with

one capacitor as shown on the below image:

Where $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \rightarrow C_{eq} = \frac{C_1 * C_2}{C_1 + C_2} = 171,42 \mu F$

Applying Kirchoff's loop rule to the circuit we obtain

$-\frac{q}{C_{eq}} - L \frac{d i}{d t} = 0$ , but $i = \frac{d q}{d t}$ therefore $\frac{d^2 q}{d t^2} = - \frac{1}{L C_{eq}} q$

The solution of the diferential equation is an armonic function

$q(t) = Q_{max} cos(\omega t + \phi)$ , where $\omega = \sqrt{\frac{1}{LC_{eq}}}$ .

SInce $i = \frac{d q}{d t}$ , then $i (t) = - I_{max} sin(\omega t + \phi)$ , where $I_{max} = \omega Q_{max}$ .

The circuit resonant frecquency is $\omega = \sqrt{\frac{1}{L C_{eq}}} = 165.88 \frac{1}{s}$ .

Applying this initial condition to

we obtain $\phi = \frac{3 \pi}{2}, \, and \, Q_{max} = \frac{I_{max}}{\omega}$

The charge in the capacitor $C_{eq}$ in is

$q(22.1 ms) = \frac{135*10^{-3} A}{165.88 s^{-1}} cos[(165.88 s^{-1}*22.1*10^{-3}) + \frac{3 \pi}{2}]$

Since are in parallel

The potential difference between b and c is $V_b - V_c = \frac{q}{C_{eq}} = -\frac{ 4.07*10^{-4} \, C}{171,42*10^{-6} F} = - 2.37 V$

To find the capacitor $C_1$ maximum charge we consider that:

$Q_{eq \, max} = Q_{1 \, max} = Q_{2 \, max} = \frac{I_{max}}{\omega} = \frac{135 *10^{-3} A}{165.88 s^{-1}} = 8.14*10^{-4} C$