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).
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