The transfer function H(s) of the given circuit is:

Transfer Function Analysis of the Given Circuit:

The transfer function H(s) of an electrical circuit represents the relationship between the input and output signals in the Laplace domain. It is expressed as a ratio of the Laplace transform of the output signal to the Laplace transform of the input signal. In this problem, we aim to determine the correct transfer function of the given circuit.

Correct Option:

The correct transfer function is:

Option 2: $\rm H(s) = \frac{R}{s^2 L R C + s L + R}$

Derivation:

To derive the transfer function, we analyze the circuit using fundamental principles such as Kirchhoff's Voltage Law (KVL) and Laplace transform techniques.

Step 1: Circuit Configuration and Parameters

The circuit under consideration is likely an RLC circuit, comprising a resistor (R), inductor (L), and capacitor (C). The configuration of the circuit determines how these components interact, influencing the transfer function. Based on the mathematical representation in the options, we assume it is a series RLC circuit.

Step 2: Applying Kirchhoff's Voltage Law (KVL)

For a series RLC circuit, KVL states that the sum of voltage drops across the resistor, inductor, and capacitor equals the applied input voltage:

$V_{in}(t) = V_R(t) + V_L(t) + V_C(t)$

Using the Laplace transform, the voltage drops can be expressed as:

• $V_R(s) = R I(s)$
• $V_L(s) = s L I(s)$
• $V_C(s) = \frac{I(s)}{s C}$

Substituting these into the KVL equation:

$V_{in}(s) = R I(s) + s L I(s) + \frac{I(s)}{s C}$

Factorizing $I(s)$ :

$V_{in}(s) = I(s) \left( R + s L + \frac{1}{s C} \right)$

$I(s) = \frac{V_{in}(s)}{R + s L + \frac{1}{s C}}$

Step 3: Output Voltage Relation

The output voltage is typically taken across one of the components. Based on the options provided, the output voltage is likely across the resistor. Therefore:

$V_{out}(s) = V_R(s) = R I(s)$

Substituting $I(s)$ into $V_{out}(s)$ :

$V_{out}(s) = R \cdot \frac{V_{in}(s)}{R + s L + \frac{1}{s C}}$

Thus, the transfer function $H(s)$ becomes:

$H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{R}{R + s L + \frac{1}{s C}}$

Simplify the denominator:

$H(s) = \frac{R}{s^2 L R C + s L + R}$

This matches Option 2, confirming its correctness.

Important Information:

To further analyze the other options, let’s evaluate their mathematical expressions:

Option 1: $\rm H(s) = \frac{C}{s^2 L + s L R C + R}$

This option incorrectly places $C$ in the numerator instead of $R$ . Moreover, the denominator does not match the standard form derived from the circuit analysis. Therefore, it is incorrect.

Option 3: $\rm H(s) = \frac{R}{s^2 L + s L R C + R}$

Although this option has $R$ in the numerator, the denominator differs from the correct one derived earlier. Specifically, it omits the term $s^2 L R C$ , making it inconsistent with the circuit's transfer function.

Option 4: $\rm H(s) = \frac{C}{s^2 L R C + s L + R}$

This option swaps $R$ for $C$ in the numerator. However, the denominator matches the correct transfer function. Since the numerator should be $R$ , this option is incorrect.

Conclusion:

The correct transfer function for the given circuit is $\rm H(s) = \frac{R}{s^2 L R C + s L + R}$ , corresponding to Option 2. This result is obtained through a systematic application of circuit analysis principles, including KVL and Laplace transform techniques.