I was absent due to the flu and I\'m lost on what we\'re supposed to do in this

Question

I was absent due to the flu and I'm lost on what we're supposed to do in this chapter

For multiple-choice questions, circle the letter of the one best answer (unless more than one answer is asked for). Show your work on free-response questions. Be sure to use proper units and significant figures in your final answers. Organization and neatness count, too. Do NOT use red ink - please reserve that color for grading. 1. You have a basic "R-C" circuit, such as the one shown here. Let R = 4.00 k Ohm. C = 750.mu F. and S=200.0 V. The capacitor is initially uncharged. At time t = 0. the switch is closed and the capacitor begins to charge. You do NOT need to show your work for fill-in-the-blank questions or graphing questions. What will be the potential across the capacitor, Vc, after a very long period of time? Vc(t rightarrow infinity)=___ What will be the eventual charge Qmax on the capacitor after a very long period of time? Q max = ___ How long is the time constant x for this R-C circuit? Show your work. At what time does Vc reach 75.0 V? Show your work.

Explanation / Answer

as u lost your class . let me teach you the basic concepts of RC circuits

when a ressitor , capacitor are connected are connected in series, then that circuit is called a simple RC circuit.

we choose initially a uncharged capacitor, so that when the circuit is completed using a source of emf,(V)

this is called the state the charging, where accumulation of charge across the terminals of capacitor takes.

when the circuit is just started, this is called initia state conditons:

at t=0, charge across q = 0 and total potental drop takes place at ressitor ( so current i = Imax = V/R)

here slowly the capacitor subjects to charge and this takes place till it gets completely charge.

this usually we call long time after the circuit is made. and at this point total pot drop occurs across the capacitor

and pot drop contributed by the ressitor is zero

so final conditions are: at t= T, q=Q and i = 0

intermediate stage at time of t, both capacitor and resitor contribute equall to the pot drop

so at this time we get equation usingn KVL,

total charge Q = CV                                 Q= final charge , I0= initai current , final current after long time is I=0

currrent i = dq/dt

so v-iR -q/C = 0

Q/C - Rdq/dt -q/C = 0

solving this differential equation for relation between Q and q we get

equautin gfor charge while charging mechanism as

q = Q I(1-e^-t/RC)-------------------------------------1

where RC is called timeconstant and denoted by T

so T = RC

similarly we get current variation as

i = io(e^-t/Rc)--------------------------------------------2

we can get the eqns of charge and current while discharging as

q = Qe^-t/RC -------------------------------------------------3

i = io e^-t/RC----------------------------------------------4

we can get the eqvation for Voltage as

V = Vo(1-e^t/RC) while charging-------------------------------------5

V= voe^-t/RC while discharging-----------------------------------------6

these 6 set of equations are usually needed to solve problems in RC circuit

so lets solve your problem now

a V= vo = 200 volts

b.apply Q= CV

find charge first

Q = 750 uF *200 = 0.15 C

c. T = RC = 4000*750uF = 3 secs

d. Vapply V = V0e^-t/RC

e^-t/RC = 75/200

e^-t/RC = 0.375

-t/RC = ln*0.375)

-t /RC = -.9808

t = 0.9808 * RC

t = 2.94 secs

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