Question

Answer for part b is R=247 and V=16v

50. (8 points) You set up the circuit shown below, C = 5.00 uF. At the time t = 0, you close the switch and then measure the charge q on the capacitor a function of the current i through the resistor. A graph of your results is also shown below. 60.00 50.00 q=-1.2475*1 + 79.9 R2 = 0.9999 40.00 Switch closed 30.00 20.00 10.00 60 i VA +q - 0.00 0 10 20 30 40 50 a R charge as a function of current for a series RC circuit (a) Explain why the data points when plotted this way, make a linear graph. Be specific (prove with equations]: (b) From the information on the graph, what are the values of the voltage of the battery and the resistance of the resistor? Show all work:

Answer 1

for RC series Circuit (a) charge at time it Current In v 2 = cv. [l-e- the ] - e-RC By 0 and ☺ Basarost q=cro [i-IR] q = cvo - (CR) I of Y=-matc as, 2 is an yo axis, I is on an axis Equation Y = K, -K₂ In = 1Y= -K₂ stky straight d line This is equation of straight line with negative slope and positive intercept on. y-axis slope of graph q = -1.2475 i +79.9. 2 =-CR)i + CV. Comparing (a) [Vo = 79.9 Mc V. 79. 981056 - 79.98156 Vo = 16 volt (6) CR = 1.24758/03 R = 1.2475X/63 - 5x10-6 2 2 = 24 9 J2