Now, a second resistor R2 of 3.3 105 Ω is connected in parallel to the existing resistor in the circuit, and a second capacitor C2 = 5.00 µF is connected to the existing capacitor in parallel. (a) What will be the new time constant? τ = s (b) What will be the maximum current in the circuit (leaving the battery terminal)? Imax = µA Now, we connect the second resistor R2 of 3.3 105 Ω and the second ca

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.

it would be kevlar ^^

force tends to be low and long diustance masses are high