For series RLC circuit given below in figure, choose the correct answer based on Kirchoff's voltage law from following:
\(Ri+L\frac{di}{dt}+\frac{1}{C}\int i\,dt = V(t)\)
To solve the problem using Kirchoff's Voltage Law (KVL) for a series RLC circuit, we need to consider the voltage drops across the resistor, inductor, and capacitor, as well as the input voltage source \( V(t) \).
According to KVL, the sum of the voltages around a closed loop must be equal to the applied voltage \( V(t) \). Therefore, the equation for the circuit becomes:
\(Ri + L \frac{di}{dt} + \frac{1}{C} \int i \, dt = V(t)\)
This corresponds to the second option given in the list.
Conclusion: The correct answer is \(Ri + L \frac{di}{dt} + \frac{1}{C} \int i \, dt = V(t)\), as it accurately represents the application of Kirchoff's Voltage Law to the series RLC circuit.
The RLC circuit given in figure below can be solved

A. Current i(t) can be solved using KVL
B. Current i(t) can be solved using KCL
C. Current i(t) can be solved using fourier transform
D. Current i(t) can be solved using laplace transform
E. Current i(t) can be solved using Fourier series
Choose the correct answer from the options given below:
Match the following in the context of RLC series circuit :
| List - I | List - II |
| (a) Under damped | (i) \(\xi=1\) |
| (b) Critically damped | (ii) \(\xi \gt 1\) |
| (c) Quality factor | (iii) \(\dfrac{1}{2\xi}\) |
| (d) Over damped | (iv) \(\xi \lt 1\) |
Codes :
Which of the following circuits will have transients ?
1. Resistive
2. R-L
3. R-C
4. R-L-C
Which is correct ?
Consider the following statements regarding circuit elements:
1. The voltage across a capacitor cannot change instantaneously.
2. The current through an inductor cannot change instantaneously.
3. The current through a capacitor is always a continuous function.
4. The voltage across an inductor is always a continuous function.
Which of these statements are correct?
At t = 0+ an inductor with zero initial condition acts as a/an
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Name that transient which is produced when a circuit, which is originally dead, is energized.
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