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.

4 Views | Posted 6 days ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

6 days ago

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

Charge on C1 = CV            Charge on C2 = 4CVWhen connected in parallel <math> <mrow> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <mi>V</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>   <math> <mrow> <mi>Q</mi> <msub> <mrow> <mo>'</mo> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>C</mi> <mi>V</mi> </mrow> </math> <math> <mrow> <mi>Q</mi> <msub> <mrow> <mo>'</mo> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>C</mi> <mi>V</mi> </mrow> </math> <math> <mrow> <mo>? </mo> </mrow> </math> <math> <mrow> <mi>E</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1820403316"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mi>E</mi> <mo>'</mo> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>   <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mi>C</mi> <msup> <mrow> <mi>V</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>5</mn> <mn>0</mn> <mi>E</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math> x = 50

0 Upvotes 0 Downvotes

Similar Questions for you

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

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

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

Five metallic plates each of area A and separated from adjacent plate by a distance d as shown in figure. The capacitance of the system across A and B is:

alok kumar singh

C = 3Aε₀/d

Two infinitely long parallel wires having linear charge densities λ₁ and λ₂ respectively are placed at a distance of R. The force per unit length on either wire will be

alok kumar singh