After having been open for a long time, the switch in the figure below is closed
ID: 1272288 • Letter: A
Question
After having been open for a long time, the switch in the figure below is closed at time t = 0. Use the following values of resistance and capacitance: R1 = 13.0 kohm, R2 = 19.0 kohm, R3 = 4.00 kohm, and C = 14.0 uF.
(a) Just after the switch is closed, what are the absolute values of the currents through the three resistors? Give your answers in microamps. through resistor 1: uA
through resistor 2: uA
through resistor 3: uA
(b) I got this answer
(c) After the switch has been closed for a long time, what are the absolute values of the currents through the three resistors? Give your answers in microamps. through resistor 1: uA
through resistor 2: uA
through resistor 3: Correct: Your answer is correct. uA
(d) After the switch has been closed for a long time, what is the charge separated by the capacitor? Give your answer in microcoulombs. uC
(e) With the switch closed, what is the time constant for the growth of charge separated by the capacitor? HINT: Use your knowledge of the maximum charging current and the maximum separated charge; to find the differential equation for the growth of charge requires using the Loop and Junction rules and solving three equations in three unknowns, which is an unnecessarily difficult strategy. s
(f) When the switch has been closed for 1.2305 10-1 s, what are the absolute values of the currents through the three resistors, and what is the charge separated by the capacitor? Give your answers in microamps and microcoulombs. HINT: Find the separated charge and the current through resistor 3 first, then apply the Loop and Junction rules. the separated charge: uC
the current through resistor 1: uA
the current through resistor 2: uA
the current through resistor 3: uA
(g) When the switch has been closed for 1.2305 10-1 s, find the power values requested below: the rate at which the ideal battery is supplying energy to the circuit: W the rate at which energy is being stored in the capacitor: W the rate at which thermal energy is being produced in resistor 1: W
the rate at which thermal energy is being produced in resistor 2: W
the rate at which thermal energy is being produced in resistor 3: W
After having been closed a long time, the clock is reset to t = 0, and the switch is suddenly reopened.
(h) Just after the switch is reopened, what are the absolute values of the currents through the three resistors? Give your answers in microamps. the current through resistor 1: uA
the current through resistor 2: uA
the current through resistor 3: uA
And what is the direction of the current through resistor 3? towards the bottom of the page, the current is zero, or towards the top of the page
(i) Just after the switch is reopened, at what rate is energy being released from the capacitor? W
(j) After the switch has been reopened, what is the time constant for decay of charge separated by the capacitor? s
(k) After the switch has been reopened for a time of 2.0930 10-1 s, find the amount of charge separated by the capacitor and the absolute value of the current which is discharging the capacitor. Give your answers in microcoulombs and microamps. charge separated by the capacitor: uC
absolute value of the discharging current: uA
(l) After the switch has been reopened for a time of 2.0930 10-1 s, find the power values requested below: the rate at which the capacitor is releasing stored energy: W
the rate at which thermal energy is being produced in resistor 2: W
the rate at which thermal energy is being produced in resistor 3: W
(m) After the switch has been reopened, find the time interval required for the charge separated by the capacitor to fall to 19.0% of its maximum value. s
Explanation / Answer
a) initially capacitor acts short ckt.
Rnet = R1 + R2*R3/(R2+R3) = 13 + 19*4/(19+4) = 16.13 kOhm
I1 = V/Rnet = 9/16.13*10^3 = 558 micro A
I2 = I1*(R3/(R2+R3) = 558*4/(19+4) = 97 micro A
I3 = I1*(R2/(R2+R3) = 558*19/(19+4) = 461 micro A
b)
c) after a long time capacitor acts as open ckt.
Rnet = R1 + R2 = 13 + 19 = 32 kohms
I1 = V/Rnet = 9/32*10^3 = 281 micro A
I2 = I1 = 281 micro A
I3 = 0
d) VC = I2*R2
= 281*10^-6*19*10^3
= 5.34 volts
Q = C*Vc
= 14*10^-6*5.34
= 74.76 micro C
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