Which of the following circuits will have transients ? 1. Resistive Which is correct ?
2. R-L
3. R-C
4. R-L-C
2, 3, 4
A transient requires an element that stores energy, and a purely resistive circuit has none — so the answer is 2, 3 and 4, option 4.
Why a resistive circuit cannot have a transient. Ohm's law is algebraic:
\(i=\dfrac{v}{R}\)
The current at any instant depends only on the voltage at that instant. There is no derivative, so there is no differential equation to solve, no time constant, and no memory of what came before. Apply a step and the current steps with it — the new state is reached immediately.
Why the other three do have transients. Each contains at least one storage element whose defining relation involves a derivative:
| Element | Relation | Cannot change instantly |
|---|---|---|
| Inductor | \(v=L\dfrac{di}{dt}\) | Its current |
| Capacitor | \(i=C\dfrac{dv}{dt}\) | Its voltage |
A discontinuous change would demand infinite voltage or infinite current. Energy stored as \(\frac{1}{2}Li^{2}\) or \(\frac{1}{2}Cv^{2}\) must be moved at a finite rate, and the interval during which it is redistributed is the transient.
The order of the circuit sets the character of the response.
First order — R-L or R-C, one storage element — gives a simple exponential with a single time constant, \(\tau=L/R\) or \(RC\). It approaches the final value monotonically and can never overshoot.
Second order — R-L-C, two storage elements — allows energy to be exchanged between L and C, so the response can oscillate. Its character depends on the damping:
\(\zeta=\dfrac{R}{2}\sqrt{\dfrac{C}{L}}\qquad\text{(series R-L-C)}\)
giving an overdamped, critically damped or underdamped approach — the last with ringing at
\(\omega_{d}=\omega_{n}\sqrt{1-\zeta^{2}}\)
Where this matters practically : stray inductance and capacitance are unavoidable, so every real "resistive" circuit shows a small transient at high enough speed — which is why fast digital edges ring on long traces.
Hence, the circuits with transients are 2, 3 and 4.
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