1a) If you only have the current through the circuit at two different times, how

Question

1a) If you only have the current through the circuit at two different times, how would you find the time constant of the RC circuit analytically? Show full work in the space below:

1b)Using the data from the graph, what is the time constant of the circuit? Show all work.

1c) Using the time constant and the value of the capacitor network, what should the resistance of the resistor + wire be? Show full work below and circle answer with proper units:

1d) With this resistance, what is the maximum current you would expect in an RC circuit with an EMF of 5.2V with this resistor? What is the maximum current you measured? What is the % difference between these two values? IS this number high or low in your opinion?

1e) From your data and graph, what is the maximum electric potential you measured across the capacitor? What should it be? Calculate the % difference. Is this reasonable?

(We used 2 (4 farad) capacitors (in series) and 245 ohms of resistors (in series.)

here is our data

Time 0.033 0.031 0.03 0.03 0.029 0.028 0.028 0.027 0.027 0.026 0.026 0.025 0.025 0.024 0.024 10 4.4 4.6 4.8 4.95 5.2 5.3 5.45 5.6 5.7 5.8 110 120 130 140 150 6.05 Time vs Current Time vs Voltage 0.034 2.62+0.663 In x 0.0316e-1.95E-03x 0.032 6 0.03 4 0.028 2 0.026 0.024 100 150 100 150

Explanation / Answer

1 a.

for a RC circuit

current as a fujcntion of time is given by

I = Io(e^(-t/RC)) [ for charging ]

if we know currents at two different times

then

I1 = Io(e^(-t1/RC))

I2 = Io(e^(-t2/RC)) [ where Io is maximum current at t = 0 and R is resistcfance of the circuit and C ius the capacitance of the circuit]

hence

I1/I2 = e^(-(t1 - t2)/RC)

ln(I1/I2) = -(t1 - t2)/RC

RC = (t2 - t1)/ln(I1/I2) = time constant

1 b. for time vs current grah the equation is

I = 0.0316*e^(-1.95*10^-3 t)

comparing it with the equation

I = Ioe^(-t/RC)

time time constant for the circuit is

RC = 1/(1.95*10^-3) = 512.82 s

1 c. given two capacitros in series C = 4 F

so effective capacitance of the capacitors in series = Ceff = C/2 = 2 F

hence resistance oif the network R = RC/C = 512.82/2 = 256.41 ohm

1 d. given emf V = 5.2 V

so maximum expected current Io = V/R = 5.2/256.41 = 0.02028 A

this number is lower than the Io as seen from the equation of current vs time in the graph which is 0.09316 A

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