The switch is thrown to position 1. What will be the current in the circuit in the steady state condition ?
2 A
In the steady state an inductor is a short circuit, so the only element left limiting the current is the 5 Ω series resistor.
\(I=\dfrac{V}{R}=\dfrac{10}{5}=2\ \text{A}\)
— option 2.
Why the inductor disappears from the calculation. The voltage across an inductor is
\(v_{L}=L\dfrac{di}{dt}\)
and "steady state" means precisely that the current has stopped changing, so \(di/dt=0\) and \(v_{L}=0\). A component with no voltage across it while carrying current is behaving as a short circuit. The 1 H value therefore does not appear in the answer at all — it governs only how long the approach takes, not where it ends.
The full transient, for context. Immediately after the switch closes the inductor opposes any change of current, so \(i(0)=0\), and the current then rises exponentially:
\(i(t)=\dfrac{V}{R}\left(1-e^{-t/\tau}\right),\qquad \tau=\dfrac{L}{R}=\dfrac{1}{5}=0.2\ \text{s}\)
After one time constant the current has reached 63 % of its final value, and after five it is within 1 % — so the steady state here is effectively reached in about a second.
| Element | At t = 0+ | At steady state |
|---|---|---|
| Inductor | Open circuit | Short circuit |
| Capacitor | Short circuit | Open circuit |
Why the second 5 Ω resistor is irrelevant. It lies on the branch reached through contact 2, which the pole is not touching. Nothing connects it into the loop, so no current flows through it. Its role is to give the inductor a discharge path when the switch is later thrown to position 2 — without it, breaking the current in an inductor would generate a large \(L\,di/dt\) spike across the opening contacts. Including it in the sum, as the distractor 1 A would require, is the error the question is set to catch.
Hence, the steady-state current is 2 A.
Similar Questions
For series RLC circuit given below in figure, choose the correct answer based on Kirchoff's voltage law from following:
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-CWhich is correct ?
Important Questions from Transient Analysis
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
During discharging of a capacitor of C = 100 µF through a resistance of 1 KΩ applied with 50 V, the voltage at the time of the it's time constant is
Name that transient which is produced when a circuit, which is originally dead, is energized.
What is the value of current at t = 5T instant in an RC network fed with voltage V where T is time constant?
