Question

An inductor of inductance
$\bigg[\rm L=\dfrac{1}{100\pi}\:H\bigg]$
a capacitor of capacitance
$\bigg[\rm C=\dfrac{1}{500\pi}\:H\bigg]$
and a resistor of resistance 2 Ω are connected in series with an AC voltage source as shown in the figure.

The voltage of the AC source is given as V = 10cos(100π t)volt.
What will be the potential difference between A and B ?

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Answer 1

Answer:

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Given Information,

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

• 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:

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

Calculating the Phase Angle we get:

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

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

Answer 2

Answer:

Given :-

• An inductor of inductance

$(L) = \frac{1}{100\pi} H$

$(C) = \frac{1}{500\pi} H$

and a resistor of resistance 2 omega are connected in series with an AC voltage source as shownin the figure .The voltage of AC source is given by

$V = 100 \cos(100\pi t) volt$

To find :-

• What is the potential difference between A and B.

Solution :-

• Here refer the attachment for more information.
• It is better to understand.

Used formulae :-

• First here we calculate XL and XC.
• $XL = w \times L$
• $XC = \frac{1}{w \times c}$

Which us called inductive resistance and capacitive resistance.

Next ,

• we calculate impendence which is

• $Z = \sqrt{ {r}^{2} + (XL - XC) {}^{2} }$

• At last we considered PD across AB .which can be written as

• V=1×(XC-XL).
• Then we get the final result ..

Hope it helps u mate .

Thank you .