Problem 1: The switch in Fig.1 has been closed for a very long time. Opens at t=0s What is the charge on the capacitor? The switch is opened at t = 0s. At what time has the charge on the capacitor decreased to 10% of its initial value? 100V 6052 ww 1052 4052 2 μF 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 S2 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 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 × 10-4 s.
Problem 1: The switch in Fig.1 has been closed for a very long time. Opens at t=0s What is the charge on the capacitor? The switch is opened at t = 0s. At what time has the charge on the capacitor decreased to 10% of its initial value? 100V 6052 ww 1052 4052 2 μF 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 S2 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 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 × 10-4 s.
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Hello, I really truly need help with part A, Part B and Part C, I keep getting the wrong answer and I don't know how to do the problem is there any chance you can help me with part A, Part B, and Part C please thank you so much.
Also, can you label which one is part A, Part B and Part C. Thank you
Transcribed Image Text:Problem 1: The switch in Fig.1 has been closed for a very long time. Opens at t=0s
What is the charge on the capacitor? The switch is opened at t = 0s.
At what time has the charge on the capacitor decreased to 10% of
its initial value?
100V
6052
ww
1052
4052 2 μF
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 S2 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 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 × 10-4 s.
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