Problem 1: The switch in Fig.1 has been closed for a very long time. What is the charge on the capacitor? The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value? Opens at t-Os 100V 6052 ww 1052 40522μF a) The fact that the switch has been closed for a very long time means that the capacitor in the scheme is fully charged, and it cre- ates a physical barrier to the current. Therefore, the current through the branch of the circuit with the 10 2 resistor and the capacitor is zero (if the capacitor was in the process of charging, there would be a non-zero current in that branch carrying the charge to the capacitor). Find the current flowing through the other two resistors. FIG. 1: The scheme for Problem 1 b) Write down the Kirchhoff's loop law for the rightmost loop in the circuit (passing through 10 2 and 40 2 resistors, and the capacitor). When you move along the loop from the negative plate of the capacitor to the positive plate, you gain the potential AVC, while moving from the positive to the negative plate you lose the potential. The signs of the charges on the plates are not provided in the scheme, but you can figure them out from the Kirchhoff's loop law itself. Make an initial guess, and if you get a negative value for AVC, it means that the actual orientation of the plates is opposite to what you have guessed. From the potential difference and the capacitance, compute the charge that is stored on the capacitor. c) After the switch is opened, you are working with an RC circuit. What is the equivalent resistance in this RC circuit? Compute the time at which the charge on the capacitor decreases to 10% of its initial value. Answer: 2.3 × 104 s.