Arrange the time constant in descending order, for the circuit given below (having different combinations of Vsupply and Capacitor (C)) (A) Vsupply = 250 V ; C = 3 µF Choose the most appropriate answer from the options given below :
(B) Vsupply = 150 V ; C = 1 µF
(C) Vsupply = 100 V ; C = 2 µF
(D) Vsupply = 250 V ; C = 6 µF
(E) Vsupply = 80 V ; C = 0.5 µF
(D), (A), (C), (B), (E)
The whole question turns on one observation: the time constant does not depend on the supply voltage at all.
\(\tau=R_{th}C\)
where \(R_{th}\) is the resistance seen by the capacitor with the source deactivated. Since \(R_{th}\) is the same in all five cases, the ordering is decided by C alone — and the various supply voltages are there purely as a distraction.
Step 1 — find the resistance seen by C. Short the voltage source. Then:
The 30 kΩ now runs from the first node to the return rail, in parallel with the 70 kΩ:
\(30\parallel70=\dfrac{30\times70}{100}=21\ \text{k}\Omega\)
Adding the 9 kΩ in series gives 30 kΩ presented to the second node, which is in parallel with the 20 kΩ there:
\(30\parallel20=\dfrac{30\times20}{50}=12\ \text{k}\Omega\)
And the 8 kΩ sits in series with the capacitor itself:
\(R_{th}=8+12=20\ \text{k}\Omega\)
Step 2 — compute the five time constants.
| Case | C | τ = 20 kΩ × C | Rank |
|---|---|---|---|
| (D) | 6 µF | 120 ms | 1st |
| (A) | 3 µF | 60 ms | 2nd |
| (C) | 2 µF | 40 ms | 3rd |
| (B) | 1 µF | 20 ms | 4th |
| (E) | 0.5 µF | 10 ms | 5th |
Descending order is therefore (D), (A), (C), (B), (E) — option 4, which is simply the capacitances in descending order.
Why the supply voltage cannot matter. The time constant describes how fast the exponential approaches its final value, not what that value is. Doubling the supply doubles the final voltage and doubles the initial charging current in the same proportion, so the shape of
\(v(t)=V_{f}\left(1-e^{-t/\tau}\right)\)
is unchanged — only its vertical scale. Case (A) and case (D) share the same 250 V supply yet differ by a factor of two in \(\tau\), which makes the point directly.
A shortcut worth noting : once it is seen that \(R_{th}\) is common to all five, the network need not be reduced at all — ranking the capacitances is enough.
Hence, the descending order is (D), (A), (C), (B), (E).
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?
