Physics Alternating Current Alternating Current Overview Class 12th Physics

The equation of current in a purely inductive circuit is 5sin(49pt – 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: ${L e t π = \frac{22}{7}}$

Option 1 - <p>1.47 <span class="mathml" contenteditable="false"> <math> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <mn>4</mn> <mn>9</mn> <mi>π</mi> <mi>t</mi> <mo>−</mo> <mn>3</mn> <mn>0</mn> <mo>°</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>1</mn> <mo>.</mo> <mn>4</mn> <mn>7</mn> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <mn>4</mn> <mn>9</mn> <mi>π</mi> <mi>t</mi> <mo>+</mo> <mn>6</mn> <mn>0</mn> <mo>°</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mn>3</mn> <mo>.</mo> <mn>1</mn> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <mn>4</mn> <mn>9</mn> <mi>π</mi> <mi>t</mi> <mo>−</mo> <mn>3</mn> <mn>0</mn> <mo>°</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mn>3</mn> <mo>.</mo> <mn>1</mn> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mrow> <mo>(</mo> <mrow> <mn>4</mn> <mn>9</mn> <mi>π</mi> <mi>t</mi> <mo>+</mo> <mn>6</mn> <mn>0</mn> <mo>°</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p>

8 Views|Posted 6 months ago

Asked by Shiksha User

1 Answer

Sort by:

V

Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

$i = 5 s i n (49 π t - 30 °)$

$L = 30 m H, π = \frac{22}{7}$

$X_{L} = ω L = 49 π * 30 * 1 0^{- 3}$

$= 49 \frac{22}{7} * 30 * 1 0^{- 3}$

$V_{L} = 23.1 s i n (49 π t + 60 °)$

Upvote

Thumbs Down Icon

Similar Questions for you

Power factor of series LCR circuit at resonance is

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:

$X_{C} = \frac{1}{ω C}$

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

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

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

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

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

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:

R = $2 Ω$

L = 2 mH

E = 9V

$i = \frac{ε}{2 R} = \frac{9 v}{4 Ω} = 2.25 A$

Just after the switch ‘S’ is closed, the inductor acts as open circuit.

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

|

1.2K

Exams

|

6.9L

Reviews

|

1.8M

Answers

Learn more about...

Physics Alternating Current 2025

Physics Alternating Current 2025

View Exam Details

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering