Voltage across parallel switches

[Metalsonic75]

The 300 micro-Farad capacitor in the figure is initially charged to 110 V , the 1200 micro-Farad capacitor is uncharged, and the switches are both open. What is the maximum voltage to which you can charge the 1200 micro-Farad capacitor by the proper closing and opening of the two switches?

I know that electric potential (aka voltage) will increase if the current decreases, and that the most rapid way to decrease current is to open a switch that has been closed for a long time. So I assume the best plan is to close switch 1, wait until the current is at its max, then close switch 2 and open switch 1. Does this sound reasonable?

I also am not sure how to calculate this transfer of current, however. Should I use the equation for potential difference deltaV = -L(dI/dt) ? I can equate L = R(tau) to get
deltaV = -LR(tau) * (dI/dt), but I am unsure where to go from there, because I am not sure how voltage is transferred between switches.

I appreciate your time.

 

[jbsteele]

Yes, your approach is reasonable. The equation you have is correct, but you need to figure out the values for L, R, and tau (the time constant). To do this, you can use the equations for capacitors in parallel and series, to calculate the total capacitance of the two capacitors in series, the total resistance between the two switches, and the time constant. Once you have these values, you can use them to calculate the maximum voltage to which you can charge the 1200 micro-Farad capacitor.

 

[Moetasim]

I can provide some insights and suggestions to help answer the question. First, let us start by understanding the concept of parallel switches and capacitors. When switches are connected in parallel, they share the same voltage across them, but the current may vary depending on the resistance of each switch. Similarly, when capacitors are connected in parallel, they share the same voltage but the capacitance adds up, resulting in a larger total capacitance.

In this scenario, when both switches are open, the 300 micro-Farad capacitor is charged to 110 V. This means that there is a potential difference of 110 V across the capacitor. The 1200 micro-Farad capacitor, on the other hand, is uncharged, meaning there is no potential difference across it.

To answer the question, we can apply the principle of conservation of charge. When the switches are closed, the total charge on the two capacitors will remain the same. This means that the charges will redistribute between the two capacitors, resulting in a new potential difference across them.

To find the maximum voltage to which the 1200 micro-Farad capacitor can be charged, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Since we know the capacitance and the initial potential difference of the 300 micro-Farad capacitor, we can calculate the initial charge on it.

Q1 = C1V1 = (300*10^-6 F)(110 V) = 33 mC

Now, when the switches are closed, the total charge will remain the same. Therefore, the charge on the 1200 micro-Farad capacitor will be the difference between the total charge and the initial charge on the 300 micro-Farad capacitor.

Q2 = Qtotal - Q1 = 33 mC

Using the equation Q = CV, we can now calculate the potential difference across the 1200 micro-Farad capacitor.

V2 = Q2/C2 = (33*10^-3 C)/(1200*10^-6 F) = 27.5 V

Therefore, by properly closing and opening the two switches, the maximum voltage to which the 1200 micro-Farad capacitor can be charged is 27.5 V.

In terms of the suggested plan, closing switch 1 and then switch 2 may not necessarily result in the maximum