Todai Entrance Exam: Subject 2013 – Problem 1

(1) Reference: 1 & 2 & 3

There are 2 ways to attack this problem. For the first way, we attack the differential equation \frac{v_{1} - v_{2}}{R} + C_{1}\frac{\mathrm{d} v_{1}}{\mathrm{d} t} = 0 subject to C_{1}v_{1} + C_{2}v_{2} = C_{1}V_{1} + C_{2}V_{2}.

For the second way, we combine the series of capacitors C_{1} and C_{2} into one capacitor C = \frac{C_{1}C_{2}}{C_{1} + C_{2}}, and solve this RC circuit with the charge conservation equation C_{1}v_{1} + C_{2}v_{2} = C_{1}V_{1} + C_{2}V_{2}.

As this is a RC circuit, as t \rightarrow \infty, v_{C} = v_{1} - v_{2} also approaches 0. Remember the charge conservation equation:

    \[ \begin{Bmatrix} C_{1}v_{1} + C_{2}v_{2} = C_{1}V_{1} + C_{2}V_{2} \\  v_{1} - v_{2} = 0 \end{matrix} \Rightarrow  v_{1} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} \]

(2) Reference: 4 & 5. Here the combined capacitor is an active component (voltage source), so i(t) will follow the sign of the current through R for convenience.

    \[ \begin{Bmatrix} v_{C} = v_{1} - v_{2} = (V_{1} - V_{2})e^{-\frac{C_{1}+C_{2}}{RC_{1}C_{2}}t} \\ C_{1}v_{1} + C_{2}v_{2} = C_{1}V_{1} + C_{2}V_{2} \end{matrix} \Rightarrow v_{1} = \frac{C_{2}(V_{1} - V_{2})}{C_{1} + C_{2}}e^{-\frac{C_{1}+C_{2}}{RC_{1}C_{2}}t} + \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} \]

    \[ i(t) = \frac{v_{C}}{R} = \frac{V_{1} - V_{2}}{R}e^{-\frac{C_{1}+C_{2}}{RC_{1}C_{2}}t} \]

(3)

    \[ E_{R}(t) = \int_{0}^{t} \frac{v_{C}^{2}(x)}{R}dx = \frac{(V_{1} - V_{2})^{2}}{R}\int_{0}^{t} e^{-2\frac{C_{1} + C_{2}}{RC_{1}C_{2}}x}dx \]

    \[ E_{R}(t) = -\frac{1}{2}\frac{C_{1}C_{2}(V_{1} - V_{2})^{2}}{C_{1} + C_{2}}e^{-2\frac{C_{1} + C_{2}}{RC_{1}C_{2}}x} \big |^{t}_{0} = \frac{1}{2}\frac{C_{1}C_{2}(V_{1} - V_{2})^{2}}{C_{1} + C_{2}}\left (1 - e^{-2\frac{C_{1} + C_{2}}{RC_{1}C_{2}}t}  \right ) \]

When t >> \frac{RC_{1}C_{2}}{C_{1} + C_{2}} (see slide 8):

    \[ E_{R} \approx \frac{1}{2}\frac{C_{1}C_{2}(V_{1} - V_{2})^{2}}{C_{1} + C_{2}} \]

The total potential energy stored in C_{1} and C_{2} at t = 0 is:

    \[ E_{C} = \frac{1}{2}Cv_{C}^{2}(0) = \frac{1}{2}\frac{C_{1}C_{2}}{C_{1} + C_{2}}(V_{1} - V_{2}) \approx E_{R} \]

(4)

Reference: A negative charge flows from the negative plate of C_{1} to the negative plate of C_{2}, which forces the positive plate of C_{2} to be more positive by having another negative charge flows from the positive plate of C_{2} to the positive plate of C_{1}, which makes the positive plate of C_{1} to be less positive. So the voltage of C_{1} keeps decreasing while the voltage of C_{2} keeps increasing, the voltage difference between the twos keeps decreasing, hence the decrease in energy (which is proportional to the voltage difference).

(5)

Thanks to Wikipedia, the equivalent circuit without non-linear elements for 0 \leq t \leq t_{0} is a battery with nonzero internal resistance r, which turns the circuit into RC circuit. Here:

    \[ \begin{Bmatrix} v(t) = V_{a} + |v_{r}(t)|\\i(t) = \frac{|v_{r}(t)|}{r}\\r = \left ( 1 - \frac{V_{a}}{V_{b}} \right )R \end{matrix} \]

(6)

    \[ v_{C}(t) = V_{a} + (V_{1} - V_{2} - V_{a})e^{-\frac{t}{rC}} \Rightarrow v_{r}(t) = V_{a} - v_{C}(t) = (V_{a} + V_{2} - V_{1})e^{-\frac{t}{rC}} \]

    \[ v_{r}(t_{0}) = V_{b} - V_{a} \Rightarrow t_{0} = \left ( 1 - \frac{V_{a}}{V_{b}} \right )RC\log\frac{V_{a} + V_{2} - V_{1}}{V_{b} - V_{a}} \]

    \[ v_{1}(t_{0}) = \frac{C_{2}(V_{1} - V_{2})}{C_{1} + C_{2}}\left ( \frac{V_{b} - V_{a}}{V_{a} + V_{2} - V_{1}} \right )^{1 - \frac{V_{a}}{V_{b}}}+ \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} \]


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