Gate Preparation

Important question on 'GATE EC' 

For any gate aspirant, the most challenging part of the preparation is revision. There are 10 subjects for EC. To deal with all those subjects you need the right strategy. During my preparation days, I followed a technique which I named "Hammering of your brain". 

How does this "hammering of your brain" technique work? Let me break it to you........... 

First, clear the basics of all the 10 subjects. Do practice previous year gate questions asked in a subject (topic-wise) while preparing for any particular subject. Now, suppose after one month or so you want to revise a particular subject. What do you do? 

You will again start reading those subjects from notes or have a look at the important parts, right? I say don't do that! In your brain, that is more volatile in nature. Eventually, after few weeks most of it will go by far in space. So what I suggest, pick a question randomly from any particular topic you want and start solving. Do that without giving a look at its background theory again (which you have already covered many times while reading that topic the first time). If you can solve that problem, that's good. If not, go check for the theory or background concept (only that part especially, nothing more!). That way your brain gets used to the unusual situation. Your brain will automatically make a permanent space for that particular topic. That's a hammer I say, stretch your permanent storage of brain every day little by little.

Now without making it boring ( what was that before! LOL) let's come to today's topic. I will take a problem on network theory which will be useful for your revision.

Q)

\[V_{in}= 50 + 5\sin t, \; \; R= 50 K\Omega \; \; V_{Oavg}=7.5V\]

\[V_{in}= 100 + 10\sin t, \; \; R= 100 K\Omega \; \; V_{Oavg}=20V\]

then find out   \(V_{Oavg}\) if  \(V_{in}= 100 + 10\sin t, \; \; R= 150 K\Omega \; \; V_{Oavg}=?\)  (capacitor value is constant and high)

Solution:

 

Now assume Capacitor is short-circuited in any finite signal frequency. Let's redraw the circuit. 

So we don't need to worry about ac signal at all . That's relaxing! Also important thing is we don't need to find out the Zth (Thevenin Impdanace) rather  we shall find Rth (Thevenin's resistance). Let's make some equation and find out that.

Due to the homogeneity property of a Linear system we can write \(V_{th}=K \times V_{in}\).

Now just redraw the whole thing for the last time 

\[V_{O,avg}= K\times V_{in} \times \frac{R}{R+R_{th}}\]

Now we can get two equation

\[7.5= K\times 50 \times \frac{50}{50+R_{th}} \; \; \; (1)\]

\[20= K\times 100 \times \frac{100}{100+R_{th}} \; \; \; (2)\]

Solving above two equation you get

\[K=0.3 , \; R_{th}=50\]

Now put those value find the answer 

                           \[0.3\times 150 \times \frac{150}{150+R_{th}} = 33.75\]

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