Question

ANSWERS MUST BE CORRECT: (Detailed explanation of order of processes required. Take your time and answer it perfectly to receive credit. The most detailed answer and explanation will be awarded points).

A series RC circuit with C = 30 uF and R = 5.4 ohm has a 18 V source in it. With the capacitor initially uncharged, an open switch in the circuit is closed.

QUESTION:

What is the current in the resistor at that time?    (Express your answer using two significant figures).

                             I = ???   A  

Answer 1

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

current at initial time in RC circuit is given by I = Imax = V/R

Imax = 18/5.4

Imax= 3.3334 Amps