An emf of <math> <mn>20</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math> is applied at time <math> <mi mathvariant="normal">t</mi> <mo>=</mo> <mn>0</mn> </math> to a circuit containing in series <math> <mn>10</mn> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">H</mi> </math> inductor and <math> <mn>5</mn> <mi mathvariant="normal">Ω</mi> </math> resistor. The ratio of the currents at time <math> <mi>t</mi> <mo>=</mo> <mi mathvariant="normal">∞</mi> </math> and at <math> <mi>t</mi> <mo>=</mo> <mn>40</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> is close to: (take <math> <mfenced open="" separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>=</mo> <mn>7.389</mn> </mrow> </mrow> </mfenced> </math>

i = V r 1 - e - R t / L = 4 1 - e - 500 t At t = ∞ i 1 = 4 A at t = 40 s i 2 = 4 1 - e - 20000 = 4 1 - 1 e 2 10000 = 4 1 - 1 ( 7.389 ) 10000 i 1 i 2 is slightly greater than 1 .

Ask & Answer Question

Alternating Current Overview Physics Alternating Current Physics Class 12th

An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$

Option 1 -

1.06

Option 2 -

0.84

Option 3 -

1.46

Option 4 -

1.15

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

$\begin{array}{r} i & = \frac{V}{r} (1 - e^{- R t / L}) \\ = 4 (1 - e^{- 500 t}) \end{array}$

At $t = \infty$

$i_{1} = 4 A$

at $t = 40 s$

$i_{2} = 4 (1 - e^{- 20000})$

= 4 [1 - \frac{1}{{(e^{2})}^{10000}}]

= 4 [1 - \frac{1}{(7.389)^{10000}}]

\frac{i_{1}}{i_{2}}

is slightly greater than 1 .

<math> <mtable columnalign="right"> <mtr> <mtd> <mrow> <mi mathvariant="normal">i</mi> </mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi mathvariant="normal">V</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">r</mi> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mi mathvariant="normal">R</mi> <mi mathvariant="normal">t</mi> <mo>/</mo> <mi mathvariant="normal">L</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow></mrow> </mtd> <mtd> <mrow> <mi> </mi> <mo>=</mo> <mn>4</mn> <mfenced separators="|"> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>500</mn> <mi mathvariant="normal">t</mi> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </mrow> </mtd> </mtr> </mtable> </math> At  <math><mi mathvariant="normal">t</mi> <mo>=</mo> <mi mathvariant="normal">∞</mi></math> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">i</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>=</mo> <mn>4</mn> <mtext> </mtext> <mi mathvariant="normal">A</mi> </math> at  <math> <mi mathvariant="normal">t</mi> <mo>=</mo> <mn>40</mn> <mtext> </mtext> <mi mathvariant="normal">s</mi> </math> <math> <msub> <mrow> <mrow> <mi mathvariant="normal">i</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>=</mo> <mn>4</mn> <mfenced separators="|"> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <msup> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20000</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> <div><div><picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753781660phpxrJBzH.jpeg" alt="" width="263" height="177" loading="lazy"></picture></div><div> <math> <mo>=</mo> <mn>4</mn> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi mathvariant="normal">e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>10000</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> </div><div> <math> <mo>=</mo> <mn>4</mn> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mn>7.389</mn> <msup> <mrow> <mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>10000</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> </div><div> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">i</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">i</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math>  is slightly greater than 1 .</div></div>

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:

Vishal Baghel

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.

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

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 m H inductor and 5 Ω resistor. The ratio of the currents at time t = ∞ and at t = 40 s is close to: (take e 2 = 7.389

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

An emf of $20 V$ is applied at time $t = 0$ to a circuit containing in series $10 m H$ inductor and $5 Ω$ resistor. The ratio of the currents at time $t = \infty$ and at $t = 40 s$ is close to: (take $e^{2} = 7.389)$