Ask & Answer Question

Energy Stored in a Capacitor Physics Electrostatic Potential and Capacitance Physics Class 12th

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

Option 1 -

1.85

Option 2 -

0.95

Option 3 -

0.105

Option 4 -

9.52

4 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

V = 10 $(1 - e^{- t / R C})$

2 = 20 $(1 - e^{- t / R C})$

$\frac{1}{10} = 1 - e^{- t / R C}$

$e^{t / R C} = \frac{10}{9}$

$\frac{t}{R C} = l n (\frac{10}{9}) = 0.105$

$C = \frac{t}{R \times 0.105} = \frac{1 0^{- 6}}{10 \times 0.105} = 0.95 μ F$

V = 10 <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>−</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mo>−</mo> <mi>t</mi> <mo>/</mo> <mi>R</mi> <mi>C</mi> </mrow> </msup> </mrow> <mo>)</mo> </mrow> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815997224"> </o:OLEObject></xml><! [endif]->2 = 20 <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>−</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mo>−</mo> <mi>t</mi> <mo>/</mo> <mi>R</mi> <mi>C</mi> </mrow> </msup> </mrow> <mo>)</mo> </mrow> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815997225"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>−</mo> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mo>−</mo> <mi>t</mi> <mo>/</mo> <mi>R</mi> <mi>C</mi> </mrow> </msup> </mrow> </math> <math> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mi>t</mi> <mo>/</mo> <mi>R</mi> <mi>C</mi> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mfrac> <mrow> <mi>t</mi> </mrow> <mrow> <mi>R</mi> <mi>C</mi> </mrow> </mfrac> <mo>=</mo> <mi>l</mi> <mi>n</mi> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mn>0</mn> <mn>5</mn> </mrow> </math> <math> <mrow> <mi>C</mi> <mo>=</mo> <mfrac> <mrow> <mi>t</mi> </mrow> <mrow> <mi>R</mi> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mn>0</mn> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>6</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>1</mn> <mn>0</mn> <mn>5</mn> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>9</mn> <mn>5</mn> <mi>μ</mi> <mi>F</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815997226"> </o:OLEObject></xml><! [endif]->

0 Upvotes 0 Downvotes

Similar Questions for you

The truth table for the given logic circuit is:

Raj Pandey

A parallel plate capacitor is charged and then charging battery is disconnected. If the plates are now pulled apart with insulated handles

alok kumar singh

Charge remains same

Energy stored in circuit 1 is E. If capacitors in circuit 1 and circuit 2 are connected in parallel as shown, the energy stored becomes $\frac{x E}{6}$ , find x.

alok kumar singh

Charge on C₁ = CV

Charge on C₂ = 4CV

When connected in parallel

$V_{C} = \frac{5 V}{3}$

$Q'_{1} = \frac{5}{3} C V$ $Q'_{2} = \frac{10}{3} C V$

$?$ $E = \frac{1}{2} C V^{2}$

$E' = \frac{25}{18} C V^{2} + \frac{25}{9} C V^{2}$

$\frac{25}{6} C V^{2} = \frac{50 E}{6}$

x = 50

Potential due to electric dipole on axial position at distance r from dipole is proportional to (assume r >> length of dipole)

alok kumar singh

$| E | = \frac{2 k P}{r^{3}}$

$E = - \frac{d v}{d r}$ , $v \propto \frac{1}{r^{2}}$

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

alok kumar singh

In air $F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (k \in_{0})} \frac{q_{1} q_{2}}{r^{2}}$

$F' = \frac{F_{0}}{K}$

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
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

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

A capacitor is connected to a 20 V battery through a resistance of 10 Ω . It is found that the potential difference across the capacitor rises to 2V in 1 μ s . The capacitance of the capacitor is________ μ F . Given I n ( 1 0 9 ) = 0 . 1 0 5

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

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$