Question

A resistor of 12Ω, capacitor of reactance 14 Ω and a pure inductor of inductance 0.1H are joined in series and placed across a 200V, 50Hz AC supply.
Calculate
(a) the current in the circuit.
(b) the phase angle between the current and the voltage. (Take π=3 for the sake of calculation)

Answer 1

Answer:

• Current (I) in the circuit is 10 A
• Phase angle (Φ) is 53°

Given:

1. Resistance (R) = 12 Ω
2. Capacitance reactance (Xc) = 14 Ω
3. Inductance (L) = 0.1 H
4. AC supply:- 200 V, 50 Hz

Explanation:

$\rule{300}{1.5}$

Firstly let's find Inductive reactance, From the formula we know,

$\bigstar\;\underline{\boxed{\sf X_{L}=\omega\;L}}$

Here,

• ω Denotes Angular frequency.
• L Denotes inductance.

Substituting the values,

$\longmapsto\sf X_{L}=\bigg(2\;\pi\;f\bigg)\times 0.1\\\\\\\\\longmapsto\sf X_{L}=\bigg(2\times 3\times 50\bigg)\times 0.1\\\\\\\\\longmapsto\sf X_{L}=\bigg(300\bigg)\times 0.1\\\\\\\\\longmapsto\sf X_{L}=30\\\\\\\\\longmapsto\sf X_{L}=30\;\Omega\qquad\dots\dots\sf (1)$

Therefore, we go the value of Inductive reactance.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Let's find the value of Impedance (Z) i.e. net resistance in the L-C-R circuit.

$\longmapsto\sf Z=\sqrt{R^{2}+\bigg(X_{L}-X_{C}\bigg)^{2}}\\\\\\\\\longmapsto\sf Z=\sqrt{\bigg(12\bigg)^{2}+\bigg(30-14\bigg)^{2}}\\\\\\\\\longmapsto\sf Z=\sqrt{\bigg(144\bigg)+\bigg(16\bigg)^{2}}\\\\\\\\\longmapsto\sf Z=\sqrt{\bigg(144\bigg)+\bigg(256\bigg)}\\\\\\\\\longmapsto\sf Z=\sqrt{\bigg(400\bigg)}\\\\\\\\\longmapsto\sf Z=20\\\\\\\\\longmapsto\sf Z=20\;\Omega\qquad\dots\dots\sf (2)$

So, let's find the net Current in the circuit.

$\longmapsto\sf I=\dfrac{V}{Z}\\\\\\\\\longmapsto\sf I=\dfrac{200}{20}\\\\\\\\\longmapsto\sf I=10\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf I=10\;A}}}}$

∴ Current in the circuit is 10 Amperes.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, for the phase angle between Current and voltage we know that,

$\longmapsto\sf \tan\phi=\dfrac{X_{L}-X_{C}}{R}\\\\\\\\\longmapsto\sf \tan\phi=\dfrac{30-14}{12}\\\\\\\\\longmapsto\sf \tan\phi=\dfrac{16}{12}\\\\\\\\\longmapsto\sf \tan\phi=\dfrac{4}{3}\\\\\\\\\longmapsto\sf \phi=\tan^{-1}\Bigg(\dfrac{4}{3}\Bigg)\\\\\\\\\longmapsto\sf \phi=53\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf \phi=53^{\circ}}}}}$

∴ Phase angle (Φ) b/w I and V is 53°.

$\rule{300}{1.5}$

Answer 2

Answer:

Current (I) in the circuit is 10 A

Phase angle (Φ) is 53°

Given:

Resistance (R) = 12 Ω

Capacitance reactance (Xc) = 14 Ω

Inductance (L) = 0.1 H

AC supply:- 200 V, 50 Hz