Solution for GATE EC 2023 question on Network Theory by Dr. Shilpa Sambhi

GATE stands for Graduate Aptitude Test in Engineering. GATE stands as an significant milestone for countless aspiring engineers who are willing to seek admission to postgraduate programs or pursuing career opportunities in the field of engineering.

For clearing gate, you have to prepare subjects like Network, Electrical measurements, Engineering Mathematics, Signals and System, Electrical Machines, Power System, and many more.

You may have seen the gate 2023 paper but you are not aware of the solution. Where you will get that? Don’t worry , Dr. Shilpa Mam had explained in detail all the solution of questions which appeared in the GATE EC 2023 of the subject Network Theory.

Whether you’re a GATE aspirant or simply looking to enhance your understanding of network subjects, this video is a must-watch. Don’t miss this opportunity to learn from a seasoned professional and improve your problem-solving skills. Subscribe to our channel and stay tuned for more valuable content to help you succeed in the GATE 2023 exam

In this article , we will provide a well-acquainted solution of GATE EC 2023 Questions for the challenging subject Network Theory.

Let us discuss the question on network theory of GATE EC 2023

Q. In the circuit shown below.

The current i flowing through 200 Ω resistor is ___ mA.

Ans. In this question we have to calculate the current.

Now, we will redraw the circuit and simplify the circuit by using V=IR

Step -1 :- Applying source transformation , we get circuits as

Step-2 :- Applying KCL at node V_A

\frac{V_{A}-2}{2000}+\frac{V_{A}}{2000}-1+\frac{V_{A}+1}{1000+200}=0 V_{A}\left [ \frac{1}{2} + \frac{1}{2} +\frac{1}{1.2} \right ]=1+\frac{1}{2}-\left ( \frac{1}{1.2} \right )

V_A= 0.636 V

Step-3 :- Calculating the current

i=\frac{V_{A}+1}{1000+200} i=\frac{0.636+1}{1.2}

i= 1.36 m A

So, simply by simplifying the circuit and applying source transformation , we can solve the problem. Then by observing the circuit just think logically that whether KCL is applicable or KVL. In this way, we can solve any network related problem.

Hope, so. You have well understood the solution of the problem.