PLZ SHOW FULL WORK WITH NUMBERS I APPRECIATE IT Problem 7 old exam question: ARC

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PLZ SHOW FULL WORK WITH NUMBERS I APPRECIATE IT

Problem 7 old exam question: ARC circuit is originally set up as shown in the figure at the right. The switch is open and the capacitor is empty. The DC power supply is 12 volts. The capacitance C 20,000 kuF, and resistance R1 100 2 and Re-60 2. 12 V (a) When the switch is just clos what is the reading of the ammeter? Show your work and reasoning. [Ans: 0.32 Al the time 0. Then it is observed that voltage across the capacitor Cis given below. Using the information given both in the problem statement and graph, show two different methods to determine the time constant when the switch is closed. (fyou use the graph to get the info, you need to clearly show how to do it on the graph.) [Ans: is] Voltage (V) 10 time M5 (c) After the switch has been closed for a long time, and then it is open Find the time that it takes for the charge on the capacitor to fall to 37% of its initial value. Show all of your work for full credit. [Ans: 3.2 sl

Explanation / Answer

a) when the switch is closed initially the capacitor is short circuited.

so, Rnet = R1*R2/(R1+R2) = 100*60/(100+60) = 37.5 ohms

so, I = V/Rnet = 12/37.5 = 0.32 A

b) from the formula, T = R1*C = 100*20000*10^-6 = 2 s


c)

Time constant when the switch is opened, T = (R1+R2)*C = (100+60)*20000*10^-6 = 3.2 s


Q = Qmax*e^(-t/T)

0.37*Qmax = Qmax*e^(-t/T)


0.37 = e^(-t/T)

e^(t/T) = 1/0.37

t = T*ln(1/0.37)

t = T = 3.2 s

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