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} (from the 3rd reference link: “charge is given by \text{capacitance} \times (V_{+plate} - V_{-plate})“, when we treat C_{1} and C_{2} as a series of capacitors, the sign of v_{2} is reversed against the plate sign).

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

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

(2)

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

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

(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) & (6)

I don’t know.


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