Question

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•Answer the question which is in attachment.

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Answer 1

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Given Information,

$\text{Inductance} = \dfrac{1}{100\pi} = 3.18 \times 10^{-3} \:\:H\\\\\\\text{Capacitance} = \dfrac{1}{500\pi} = 0.63 \times 10^{-3} \:\:F\\\\\\\text{Resistance} = 2 \:\:\Omega$

V = 10 Cos ( 100πt ) volts

Calculating XL and XC we get:

⇒ XL = ω × L

⇒ XL = 100 π × 3.18 × 10⁻³

⇒ XL = 0.99582 ≈ 1 Ω

Hence the Inductive Reactance is 1 Ω.

⇒ XC = 1 / (ω × C)

⇒ XC = 1 / (100π × 0.63 × 10⁻³)

⇒ XC = 1 / ( 0.19782 ) ≈ 1 / (0.2)

⇒ XC = 5 Ω

Hence the Capacitive Reactance is 5 Ω.

Now considering the connection across AB, we notice only a capacitor and a inductor. Hence it is a L-C Circuit. But, since the value of XC > XL, the circuit behaves more similar to a capacitive circuit. Hence we perform all calculations with respect to Capacitive circuit.

We know that Impedance (Z) is calculated as:

$\implies Z = \sqrt{ R^2 + (X_{L} - X_{C})^2}\\\\\implies Z = \sqrt{ 2^2 + (1 - 5)^2}\\\\\implies \boxed{ \bf{Z = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \:\: \Omega}}$

Calculating the Phase Angle we get:

$\implies \text{Phase Angle} = tan^{-1} \left[\begin{ \dfrac{X_C - X_L}{R}}\right] \\\\\\\implies \text{Phase Angle} = tan^{-1} \left[\begin{ \dfrac{5-1}{2}}\right]\\\\\\\implies\boxed{ \text{Phase Angle} = tan^{-1} (2) = 63.4^\circ}$

Now we know that,

$\implies V = I \times Z\\\\\implies I = \dfrac{V}{Z}\\\\\\\implies I = \dfrac{10.Cos(100\pi t + 63.4^\circ)}{2 \sqrt{5}}\\\\\\\implies I = \dfrac{5}{\sqrt{5}} \times Cos(100\pi t + 63.4^\circ)\\\\\\\implies \boxed{ \bf{I = \sqrt{5}.\:Cos(100\pi t + 63.4^\circ)}}$

Now considering P.D. across AB, we get:

⇒ V = I × (XC - XL)

⇒ V = √5 Cos ( (100πt + 63.4°) - 90° )

We are subtracting 90° from phase as Voltage lags by 90° in a capacitive circuit. Hence the value of V across AB is:

⇒ V across AB = √5 Cos ( 100πt - 26.6° ) V

I learnt this sum From that moderator