GATE 2020 EE Q.45: solution and answer

GATE 2020 · EEQ.45 · 2 marks · NAT

A resistor and a capacitor are connected in series to a 10 V dc supply through a switch. The switch is closed at $t = 0$ , and the capacitor voltage is found to cross 0 V at $t = 0.4 τ$ , where $τ$ is the circuit time constant. The absolute value of percentage change required in the initial capacitor voltage if the zero crossing has to happen at $t = 0.2 τ$ is ________ (rounded off to 2 decimal places).

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Answer and solution

ANSWER

The initial capacitor voltage has to change from about $- 4.918 V$ to about $- 2.214 V$ , an absolute change of $54.98%$ .

Official answer: 54.00 to 56.00Source: Official answer key, GATE 2020 EE (IIT Delhi)

Write the first-order step response of a series RC circuit: $v_{C} (t) = v_{C} (\infty) + [v_{C} (0^{+}) - v_{C} (\infty)] e^{- t / τ}$ , with $v_{C} (\infty) = 10 V$ because the capacitor charges to the supply voltage. Let $V_{0} = v_{C} (0^{+}) = v_{C} (0^{-})$ , since capacitor voltage cannot jump.
For $t > 0$ the capacitor voltage moves steadily towards $+ 10 V$ , so it can only cross zero if it starts negative, i.e. $V_{0} < 0$ . Setting $v_{C} (t_{0}) = 0$ : $10 + (V_{0} - 10) e^{- t_{0} / τ} = 0 \Rightarrow V_{0} = 10 (1 - e^{t_{0} / τ})$ .
Present case, $t_{0} = 0.4 τ$ : $V_{0} = 10 (1 - e^{0.4}) = 10 (1 - 1.49182) = - 4.9182 V$ .
Required case, $t_{0} = 0.2 τ$ : $V_{0}^{'} = 10 (1 - e^{0.2}) = 10 (1 - 1.22140) = - 2.2140 V$ .
Percentage change relative to the original initial voltage: $\frac{V _{0}^{'} - V _{0}}{V _{0}} \times 100 = \frac{- 2.2140 - ( - 4.9182 )}{- 4.9182} \times 100 = \frac{2.7042}{- 4.9182} \times 100 = - 54.98%$ The initial voltage has to be reduced in magnitude by about 55%.
Check in closed form: $\frac{e ^{0.4} - e ^{0.2}}{e ^{0.4} - 1} = \frac{e ^{0.2} ( e ^{0.2} - 1 )}{( e ^{0.2} - 1 ) ( e ^{0.2} + 1 )} = \frac{e ^{0.2}}{e ^{0.2} + 1} = \frac{1.22140}{2.22140} = 0.54983$ , so the absolute percentage change is $54.98%$ .

KEY IDEA

A capacitor charging towards $+ 10 V$ can only pass through zero if its initial voltage is negative. Setting $v_{C} (t_{0}) = 0$ in $v_{C} (t) = 10 + (V_{0} - 10) e^{- t / τ}$ gives $V_{0} = 10 (1 - e^{t_{0} / τ})$ . Work out $V_{0}$ for both crossing times and compare them. The ratio simplifies to $e^{0.2} / (1 + e^{0.2})$ .

WATCH OUT FOR

Assuming the capacitor starts uncharged (then it never crosses 0 V), or putting $e^{- t_{0} / τ}$ in place of $e^{t_{0} / τ}$ when solving for $V_{0}$ . Another mistake is dividing the change by the new voltage $V_{0}^{'}$ instead of the original $V_{0}$ , which gives about 122% instead of 54.98%.

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