Most Important Interview Questions on RC Circuit

In this tutorial most important interview questions on RC circuits will be covered.

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In this tutorial, we shall discuss questions on RC circuits which will be asked if you are appearing for an interview for the analog profile. It's asked to the candidates who are appearing for digital domain too. It's equally important for the ISRO, BARC, DRDO & ESE interviews.

If you observe the questions asked by the interviewer carefully have a pattern. The interviewer wants to hire the best candidate suitable for the profile. An interview continues on an average of 1-2 hours. But it's impossible for them to find out the best candidate for the role within that short span of time. So they ask a set of questions to judge your ability to think. So if you are able to incline your thoughts in that track, you have the upper hand in the interview.

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Important points to remember:

1)  Break your circuits into a number of nodes. Find out equivalent impedances in between those nodes. Try to do all the things in your mind only.

2) Check for the resistor connected to the node of our interest. It will help you to find out the terminal potential.

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Now let's jump into the questions keeping the exact pattern set by an interviewer to judge your ability to think.

Always try to solve by yourself by your own method. Then check for the solution.

Interviewer: Draw a circuit of two capacitors in series and apply unit-step voltage to one end and take the output from the middle.

Your mind: easy! I can do that very easily.

Voltage gets divided by the rule of a capacitive divider. No resistor connected, output will be reflected instantaneously. 

\( V_{out} = V_{in} \times  \frac{C_{1}}{C_{1}+C_{2}} \)

 

Result: 

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After this interviewer can ask for two things. I will list out both cases.

Case1

Interviewer: Connect a resistor in parallel with C2 and find the response again.

Your mind: Hold on a second, I can do that! But..........

Wait, is it a 1st order or a 2nd order circuit? I need to check the degrees of freedom! hmm....

Say initial voltage across \(C_{1}\) and \(C_{2}\) are \(V_{1}\) and  \(V_{2}\) respectively.

                                     \[V_{in}= V_{C2}+V_{C2}\]

So the degree of freedom is 1. Wow! 

Now the job is easy.Now important nodes in the circuits are \(V_{in}\) and \(V_{out}\). At t=0 unit step changes its value instantaneously, which can be thought of as high frequency. For high-frequency capacitor acts as a short circuit.

Schematic at t=0 

At t=0 output potential  

                   \[V_{out}= V_{in} \times  \frac{C_{1}}{C_{1}+C_{2}}\]

At steady state

At steady-state Vout is disconnected from Vin and Resistor R2 will pull down the Vout node to the ground.

Result:

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Now come to case 2 for 2nd question.

Case 2

Interviewer: Connect a resistor in parallel with C1 and find the response again.

Your mind: Is it inception or what! I know the trick already (LOL).

In the same manner, you say that the circuit is of order 1. At t=0 Vout potential is found by the capacitor divider rule. At steady-state, no current flows through R1, and the Vout node is shorted to the Vin node.

 Result:

Result

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After successfully answering the above two questions, you will be expecting one more question. 

Interviewer: Connect one more resistor in parallel to other capacitor.

Your mind: After solving the last two questions, I can solve this one too.

At t=0, the capacitors are dominant (as they act as a short circuit and I told you earlier to take the smaller one if they are in parallel).

\[V_{out}= V_{in} \times  \frac{C_{1}}{C_{1}+C_{2}}\]

At steady-state, the capacitors are open. Resistors will decide the Vout node potential.

                           \[V_{out}= V_{in} \times  \frac{R_{2}}{R_{1}+R_{2}}\]

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case I: R1 C1 < R2 C2

R1C1< R2C2 implies, C1/(C1+C2) < R2/(R1+R2)

Initial value is less than final value.

Result:

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Case II: R1 C1 = R2 C2

R1C1= R2C2 implies, C1/(C1+C2) = R2/(R1+R2)

The initial value is equal to the final value.

Result:

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Case III: R1 C1 > R2 C2

R1C1 > R2C2 implies, C1/(C1+C2) > R2/(R1+R2)

The initial value is greater than the final value.

Result:

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Conclusions: Use the tricks of elements to consider. Also using the pole-zero concept and bode plot you can draw the transient behavior too. I will cover that part later on.

For more updates follow my BLOG. For other tutorials check INDEX. For GATE preparation.