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

Option 1 -

88 A and tan⁻¹(11/6)

Option 2 -

8.8 A and tan⁻¹(11/6)

Option 3 -

0.88 A and tan⁻¹(11/6)

Option 4 -

8.8 A and tan⁻¹(6/11)

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)

0 Upvotes 0 Downvotes

Similar Questions for you

Power factor of series LCR circuit at resonance is

Vishal Baghel

At resonance V_L = V_C

_{$\Rightarrow c o s ? = \frac{V_{R}}{\sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}} = 1$}

An electric bulb and a capacitor are connected in series with an AC source. On increasing the frequency of the source, the brightness of the bulb:

Vishal Baghel

$X_{C} = \frac{1}{ω C}$

On increasing $ω$ XC decreases so Z decreases, current in circuit increases.

A direct current of 4 A and an alternating current of 4 A flows through two identical resistors. The ratio of heat produced in the two resistances in same time interval will be

Vishal Baghel

$\frac{H_{D . C .}}{H_{A . C .}} = \frac{I^{2} R}{I_{r m s}^{2} R} = 1$

The value of alternating emf (E) in the given circuit will be

Vishal Baghel

$E = \sqrt[]{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$

$E = \sqrt[]{{(60)}^{2} + {(100 - 20)}^{2}}$

E = 100V

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be: