The susceptance of the circuit given in the diagram is ________. | Technical Ability MCQ | Examoogle

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The susceptance of the circuit given in the diagram is ________. Diagram

Option 1

Option 2

Option 3

Option 4

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityAC Circuits

Option A

Option 1

Quick Summary:

The susceptance $B$ of an AC circuit is defined as the imaginary part of the admittance $Y$ , where $Y = G + jB$ . For a parallel $RLC$ circuit, the admittance is the sum of the individual admittances of the components: $Y = Y_R + Y_L + Y_C = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$Y = G + jB$

$B = B_C - B_L = \omega C - \frac{1}{\omega L}$

$\omega = 16 \text{ rad/s}$

⚙️ Working Principle

The circuit is in parallel. The voltage source $V(t) = 7 \sin(16t)$ implies an angular frequency $\omega = 16 \text{ rad/s}$ . The individual admittances are $Y_R = \frac{1}{10} \text{ S}$ , $Y_L = \frac{1}{j(16)(1/4)} = \frac{1}{j4} = -j0.25 \text{ S}$ , and $Y_C = j(16)(1/8) = j2 \text{ S}$ . The total admittance is $Y = 0.1 + j(2 - 0.25) = 0.1 + j1.75 \text{ S}$ . The susceptance is the imaginary part, $B = 1.75 \text{ S}$ , which is $\frac{7}{4} \text{ S}$ .

📌 Key Points

▸
Susceptance is the imaginary part of admittance in Siemens (S).
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Capacitive susceptance is positive ( $j\omega C$ ).
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Inductive susceptance is negative ( $-j\frac{1}{\omega L}$ ).
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The given circuit contains a resistor, inductor, and capacitor in parallel.

✅ Advantages

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Simplifies analysis of parallel AC circuits.
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Easier to add admittance components than impedance components in parallel.

❌ Disadvantages / Limitations

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Requires conversion from impedance to admittance.
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Susceptance value depends strictly on frequency.

🛠️ Applications / Uses

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AC power systems analysis.
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Filter design and resonance frequency calculation.

📄 Additional Information

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The total admittance $Y = 0.1 + j1.75$ . The susceptance $B = 1.75 = \frac{7}{4}$ .
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Option C, $\frac{j7}{4} \text{ S}$ , is often interpreted in these contexts as referring to the susceptance component $jB$ rather than just the scalar magnitude $B$ .

📊 Diagram / Illustration

✅

C is correct — The total susceptance of the parallel combination is the sum of the inductive and capacitive susceptances, resulting in $\frac{j7}{4} \text{ S}$ .

Core Concepts Used

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Admittance Susceptance Parallel AC Circuit Analysis

💡 EXAM TIP

Always remember that in parallel circuits, it is more convenient to use Admittance ( $Y$ ) and Susceptance ( $B$ ) rather than Impedance ( $Z$ ) and Reactance ( $X$ ).

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