Question

Given is the circuit diagram of a simple RC circuit. In the given circuit the value of capacitor ‘C’ is 100uF, what should be the value of resistor ‘R’ so that the time constant of this circuit is 15 seconds. Implement this circuit on Proteus; you have to find voltage across capacitor ‘Vc’ (Both theoretically and through simulations) at different intervals of time after the switch has been closed as mentioned in the table given below. Vs=10V

Switch R ماما Vs + C

                                                              

-Solve with all the steps, even the most simplest steps. All steps are needed.

-Also fill table with all steps.

-All steps are must.

-Will rate positive. :)

-Please be super neat and all words should be precise and clear.

Answer 1

Solution:

Applying Kirchoff's law in the loop, we can write that

VS = VR + VC ............(1), where

VS = Supply voltage

VR = Voltage across resistor

VC = Voltage across capacitor

Now, we can write the above equation as

VS = IR + q/C....(2), where

I = current

q = charge in the capacitor

Again eq (2) can be written as

VS = dq/dt R + q/C [As current is rate of change of charge]

On rearranging the above equation, we can write that

-RC dq/dt = q - CV

Or, dq / (q - CV) = -dt / RC

Now, integrating the above equation by putting the appropriate limits, we have

$\int_{0}^{q}$dq / (q - CV) = -dt/RC

07

Or, ln (q - CV) / -CV = -t / RC

On eliminating the logarithm, we get

(q - CV) / -CV = e-t / RC

Or, (q - CV) = - CV e-t / RC

Or, q = CV - CVe-t / RC

Or, q = CV (1-e-t / RC)........(3)

Now, we know that

I = dq/dt

Thus, ondifferentiating eq(3), we have

I = d/dt (CV (1-e-t / RC)

Or, I = d/dt (CV - CVe-t / RC)

Or, I = d/dt (CV) - d/dt (CVe-t / RC)

Or, I = 0 - CVe-t / RC * (-1/RC)

Or, I = V/R e-t / RC .....(4)

Thus, voltage across resistor R = VR = IR = V/R e-t / RC * R

Or, VR = VSe-t / RC

Now from eq(1), we have

VC = VS - VR

Or, VC = VS -  VSe-t / RC

Or, VC = VS (1 - e-t / RC )........(5)

Thus, while charging, the voltage across the capacitor will be given by eq (5)

Now, the time constant of an RC circuit is the time taken by the capacitor to charge from 0 to 63.2% of the applied voltage and is given by

= RC.......(6)

7

Here,

= 15 s

7

C = 100 X 10-6 F

Thus, from eq (6), we have

R = / C

7

Or, R = 15/100 X 10-6

Or, R = 150 Kohm

Now, we need to find the voltage across capacitor at t = 1
7
, 2
7
, 3
7
, 4
7
, 5
7

From eq(5), we have

VC = VS (1 - e-t / RC )

At t = 1
7

VC = 10 (1 - e-RC / RC ) V [As
7
= RC]

VC = 10 (1 - e-1)

Or, VC = 10(1 - 2.718)

Or, VC = 10(-1.718)V = -17.18 V

At t = 2
7

VC = 10 (1 - e-2RC / RC ) V

Or, VC = 10 (1 - 0.135)

Or, VC = 10 * 0.865 = 8.65 V

At t = 3
7

VC = 10 (1 - e-3RC / RC ) V

Or, VC = 10 (1 - 0.049) V

Or, VC = 10 * 0.951 V = 9.51 V

At t = 4
7

VC = 10 (1 - e-4RC / RC ) V

Or, VC = 10 (1 - 0.018) V

Or, VC = 10 * 0.982 V = 9.82 V

At t = 5
7

VC = 10 (1 - e-5RC / RC ) V

Or, VC = 10 (1 - 0.0067) V

Or, VC = 10 * 0.993 V = 9.93 V