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The transfer function $\frac{V_2(s)}{V_1(s)}$ of the circuit shown below is [Series capacitor $100\,\mu$F from input to output node; from output node to ground a series branch of $10\,\text{k}\Omega$ resistor and $100\,\mu$F capacitor; output taken across the output node to ground.]

1. $0.5\,\frac{s}{s+1}$
2. $3\,\frac{s}{s+2}$
3. $\frac{s}{s+1}$
4. $\frac{s}{s+2}$

Correct answer: $\frac{s}{s+2}$

Solution

Using impedances, the series capacitor and the load branch form a voltage divider. After simplification with the given values, the transfer function reduces to $\frac{s}{s+2}$.

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