GATE 2014 EC (EC2) – Question 42

Networks, Signals and Systems · Time and Frequency Domain Analysis of Linear Circuits · 2 marks · Multiple choice

The capacitor shown is initially uncharged. Find branch current I(t) (mA) for t > 0.

Diagram for GATE 2014 EC (EC2) question 42

$I(t)=\frac53(1-e^{-t/\tau}),\ \tau=\frac23$ ms
$I(t)=\frac52(1-e^{-t/\tau}),\ \tau=\frac23$ ms
$I(t)=\frac53(1-e^{-t/\tau}),\ \tau=3$ ms
$I(t)=\frac52(1-e^{-t/\tau}),\ \tau=3$ ms

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Correct answer: (A) $I(t)=\frac53(1-e^{-t/\tau}),\ \tau=\frac23$ ms

Explanation

The final capacitor voltage is the divider value5×2/3=10/3V, so the final2kΩ current is5/3mA. The capacitor sees Rth=1kΩ∥2kΩ=2/3kΩ, giving τ=RthC=2/3ms. Starting uncharged, **I(t)=(5/3)(1−e^(−t/τ))mA,A**.