A 0.07 H inductor and a 12 \Omega resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take \pi as 22 \over 7 ]

8.8 A and \tan ^ - 1( 11 \over 6 )

88 A and \tan ^ - 1( 11 \over 6 )

0.88 A and \tan ^ - 1( 11 \over 6 )

8.8 A and \tan ^ - 1( 6 \over 11 )

Alternating Current PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q63 with Solution

View written solutionFree

Correct answer: A

Given data

Inductance: $L = 0.07\,\text{H}$
Resistance: $R = 12\,\Omega$
AC voltage: $V = 220\,\text{V}$
Frequency: $f = 50\,\text{Hz}$
Take $\pi = \dfrac{22}{7}$

Calculate inductive reactance

Inductive reactance is

X_L = 2\pi f L

Substituting:

X_L = 2\times \frac{22}{7}\times 50\times 0.07

Since $0.07 = \frac{7}{100}$ ,

X_L = 2\times \frac{22}{7}\times 50\times \frac{7}{100}

Now simplify:

X_L = 2\times 22\times \frac{50}{100} = 44\times \frac{1}{2} = 22\,\Omega

Find impedance of the series RL circuit

For a series $RL$ circuit,

Z = \sqrt{R^2 + X_L^2}

So,

Z = \sqrt{12^2 + 22^2} = \sqrt{144 + 484} = \sqrt{628}

Z \approx 25.06\,\Omega

Calculate current

Using Ohm’s law for AC,

I = \frac{V}{Z}

Thus,

I = \frac{220}{25.06} \approx 8.8\,\text{A}

Find phase angle

For a series $RL$ circuit,

\tan \phi = \frac{X_L}{R}

Hence,

\tan \phi = \frac{22}{12} = \frac{11}{6}

Therefore,

\phi = \tan^{-1}\left(\frac{11}{6}\right)

This is the phase angle by which voltage leads current (or current lags voltage).

Check options

A: $8.8\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ✅
B: $88\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ❌
C: $0.88\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ❌
D: $8.8\,\text{A}$ and $\tan^{-1}\left(\frac{6}{11}\right)$ ❌

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

More from Alternating Current

View written solutionFree

Correct answer: A

Given data

Inductance: $L = 0.07\,\text{H}$
Resistance: $R = 12\,\Omega$
AC voltage: $V = 220\,\text{V}$
Frequency: $f = 50\,\text{Hz}$
Take $\pi = \dfrac{22}{7}$

Calculate inductive reactance

Inductive reactance is

X_L = 2\pi f L

Substituting:

X_L = 2\times \frac{22}{7}\times 50\times 0.07

Since $0.07 = \frac{7}{100}$ ,

X_L = 2\times \frac{22}{7}\times 50\times \frac{7}{100}

Now simplify:

X_L = 2\times 22\times \frac{50}{100} = 44\times \frac{1}{2} = 22\,\Omega

Find impedance of the series RL circuit

For a series $RL$ circuit,

Z = \sqrt{R^2 + X_L^2}

So,

Z = \sqrt{12^2 + 22^2} = \sqrt{144 + 484} = \sqrt{628}

Z \approx 25.06\,\Omega

Calculate current

Using Ohm’s law for AC,

I = \frac{V}{Z}

Thus,

I = \frac{220}{25.06} \approx 8.8\,\text{A}

Find phase angle

For a series $RL$ circuit,

\tan \phi = \frac{X_L}{R}

Hence,

\tan \phi = \frac{22}{12} = \frac{11}{6}

Therefore,

\phi = \tan^{-1}\left(\frac{11}{6}\right)

This is the phase angle by which voltage leads current (or current lags voltage).

Check options

A: $8.8\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ✅
B: $88\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ❌
C: $0.88\,\text{A}$ and $\tan^{-1}\left(\frac{11}{6}\right)$ ❌
D: $8.8\,\text{A}$ and $\tan^{-1}\left(\frac{6}{11}\right)$ ❌

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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