Answer:
Incomplete questions
Check attachment for the first aspect of the question.
Explanation:
So let analyse the question first.
Let the former circuit have a resistor of resistance R, if a new resistor (3.3×10^5 Ω) is connected to it in parallel to R,
Then, the equivalent resistance can be calculated as
1/Req= 1/R1 + 1/R2
Therefore,
Req = R1•R2 / (R1+R2)
Req= R×3.3×10^5 / (R +3.3×10^5)
Req= 3.3×10^5R / (R +3.3×10^5)
Also, let assume the former circuit has a capacitor of capacitance C, and a new capacitor of capacitance (5 µF) is connected in parallel to the capacitor
Then, the equivalent capacitor is
Ceq=C1+ C2
Ceq= C+ 5µF
Ceq= C + 5×10^-6 F
Now,
a. Time constant of a RC circuit is given as
τ= RC
Then, τ=Req•Ceq
τ=3.3×10^5R / (R +3.3×10^5) × (C+ 5×10^-6F)
τ=(3.3×10^5R) (C+5×10^-6)/ (R+3.3×10^5)
τ=(3.3×10^5•RC+1.65R)/(R+3.3×10^5)
b. For maximum current in an RC circuit, the maximum current occur when the the exponential function is 1 I.e, at t=0
I = Ioe^(-t/RC)
So if t=0
I=Io
So, at this point all the current appears at the resistor
Using ohms law
V=IoR
Then, Io=V/R
Io=V/Req
Io=V ÷ 3.3×10^5R / (R +3.3×10^5)
Io= V(R +3.3×10^5) / (3.3×10^5R)
Io= (VR + 3.3×10^5V)/(3.3×10^5R)
This is the analysis using any initial resistor and capacitor.
Note : the C and R are the initial resistance and capacitance of the circuit before parallel connection.
Now, using the original question, check attachment for the question.....
Now, given that the initial resistor has a resistance of 8×10^5Ω
R=8×10^5Ω
Also a capacitor of capacitance 5µF
C=5×10^-6F
And the EMF= 12V.
So, to calculate the time constant of the given question
Since we already have the function in question a
a. Time constant
τ=(3.3×10^5•RC+1.65R)/(R+3.3×10^5)
Since, R=8×10^5Ω and C=5×10^-6F
Then,
τ=(3.3×10^5RC+1.65R)/ (R+3.3×10^5)
Where RC Is the time constant of the circuit without parallel connection
τ= RC = 8×10^5×5×10^-6= 4seconds
τ=(3.3×10^5×4+1.65×8×10^5) / (8×10^5+3.3×10^5)
τ=(13.2×10^5 + 13.2×10^5)/(11.3×10^5)
τ=(26.4×10^5) / (11.3×10^5)
τ= 2.34 seconds
b. Also, for the maximum current
Let use the function got in question b
Io= (VR + 3.3×10^5V)/(3.3×10^5R)
Io= (12×8×10^5+ 3.3×10^5V)/(3.3×10^5×8×10^5)
Io= (96×10^5+3.3×10^5V) / (26.4×10^10)
Io= (99.3×10^5) / (26.4×10^10)
Io=3.76×10^-5A
Which is
Io=0.376×10^-6A
Io=0.376µA
The maximum current is 0.376µA.
You can as well change the value of the initial R and C if you have other values.
