# FreBlogg

## Solution | Question of the Day 1 | GATE - 2015 | ECE | EEE

00:29 Posted by DurgaSwaroop
In the given circuit the input is an Voltage source which has a value of 10.u(t), where u(t) is the Unity Step Function.  The Resistance of the Resistor is 1 Ohm. And, two capacitors of 3F and 2F and connected in series to this Resistor as shown in the figure. Find the voltage across the 2F capacitor as a function of time, V(t).
Assume the Capacitors are initially carrying zero charge.

Answer : V(t)=6 - 6.e^(-5/6 t)
So, That was the question and the answer. Before you proceed to solving the question, there are a few things you need to take a look at.
1. Laplace Transforms & Inverse Laplace Transforms
2. Basic Circuit relations (KVL, KCL, Voltage Division, Current Division)
3. Basics of RL, RC, RLC circuits
Assuming you have the basic knowledge of the above mentioned topics, lets proceed to the solution.
We present here two methods of solving this question. You might have used a different method to get the solution. But, as long as you got the same answer your method should be fine. If you think your method is much simpler, share it with us and we'll let others know about it.

Method 1:

We can easily transform this circuit in to its Laplace Domain Equivalent and it will be very easy to work in that domain. After we got the required quantity in S-domain, we'll take Inverse Laplace Transform of it and we'll have our answer.
So, Laplace domain equivalent of the given circuit looks like this. Resistor will remain the same. Capacitor 'C' gets replaced by 1/SC. (Inductor L gets replaced by SL ). And, the voltage source of 10V gets replaced by 10/S

(If initial state is not-zero, then there will be a few more extra things to worry about, but since the initial state was uncharged, we don't need to care about those.)
One we have that , it is easy to get the voltage across the capacitor..
First we'll find the current in the circuit as Total voltage divided with the total Impedance.

I(s)=(10/s)/[1+(1/3s)+(1/2s)] = 10/[s+(5/6)]

Once we have the current, finding the voltage is easy. Just multiply with the corresponding impedance. So, Voltage across the 2F capacitor V(S) will be,
V(s)=I(s).1/2s = 5/[s (s + 5/6 )] = (6/s)  - [6/(s + 5/6) ]

Now, we have V(s). To find v(t), we just have to take the Inverse Laplace of this, which will be,.

V(t)=6 - 6.e^(-5/6 t)

So, That's it. That is the answer. Its very simple and it hardly takes 40 secs to complete this. So, That is the first way to get the solution. Now, lets see the second way.

Method 2:

This method is applicable if the circuit is a First-order circuit as the one given in the question is.
Any quantity 'A' in the circuit can be found from the following equation very easily,
All we need to know is its value at t=0- (Initial Value) and its value at t=Inf (steady state value)

A(t)=A(∞)+[A(0-)-A(∞)]e^(-t/τ) , τ - Time constant

So, for our problem 'A' is the voltage across 2F. so,

V(t)=V(∞)+[V(0-)-V(∞)]e^(-t/τ)
V(∞) is the voltage across the 2F capacitor after reaching steady-state. At, steady state the capacitor is treated as a open circuit. So, total voltage of 10v is dropped across the two capacitors. So, by voltage division principle of capacitors,

V(∞)=[3/(3+2)]∗10=6 volts

V(0-) is the voltage initially which was given as zero. so,

V(0-)=Initial state=0 volts
Now, only thing left out is the time constant. Time constant of a RC circuit is 'RC'. In our case we will take the equivalent capacitance of the two capacitors.

τ = R.C_eq= (1Ω).(2F ||3F)
=1[(2∗3)/(2+3)]
=6/5  sec
So, we now have everything we need. Just substitute them in the above equation and you'll have your value.
V(t)=6 - 6.e^(-5/6 t)

This method is relatively simple compared to the previous method. You just need to remember the simple formula and that is all there is to it. This should take around 30 secs to solve.

So, those are two of the methods in which you can get the answer. If you didn't understand something in the explanation please let us know and we'll help you with that by providing more explanation at that particular point.
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As always have a Happy reading and Good luck with your preparation!
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