Question: Q. 12. Two identical capacitors of $12 \mathrm{pF}$ each are connected in series across a battery of $50 \mathrm{~V}$. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now?

Also find the charge drawn from the battery in each case.

U] [Delhi III 2017]

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Solution:

Ans. Equivalent capacitance in series

Energy in series combination $1 / 2$

Charge in series combination $1 / 2$

Equivalent capacitance in parallel combination $1 / 2$

Energy in parallel combination $1 / 2$ Charge in parallel combination $1 / 2$

In series combination :

$$ \frac{1}{C_{s}}=\left(\frac{1}{12}+\frac{1}{12}\right)(p F)^{-1} \quad 1 / 2 $$

$\therefore \quad C_{\mathrm{s}}=6 \times 10^{-12} \mathrm{~F}$

$U_{\mathrm{s}}=\frac{1}{2} \mathrm{CV}^{2}$

$U_{\mathrm{s}}=\frac{1}{2} \times 6 \times 10^{-12} \times 50 \times 50 \mathrm{~J}$

$$ \begin{aligned} U_{\mathrm{s}} & =75 \times 10^{-10} \mathrm{~J} \ q_{\mathrm{s}} & =C_{\mathrm{s}} V \end{aligned} $$

$$ =6 \times 10^{-12} \times 50 \mathrm{C} $$

$$ \begin{aligned} & =300 \times 10^{-12} \mathrm{C} \ & =3 \times 10^{-10} \mathrm{C} \end{aligned} $$

$1 / 2$

In parallel combination :

$$ \begin{aligned} C_{p} & =(12+12) p \mathrm{~F} \ C_{p} & =24 \times 10^{-12} \mathrm{~F} \ U_{p} & =\frac{1}{2} \times 24 \times 10^{-12} \times 2500 \mathrm{~J} \ U_{p} & =3 \times 10^{-8} \mathrm{~J} \ q_{p} & =C_{p} V \ & =24 \times 10^{-12} \times 50 \mathrm{C} \ & =1.2 \times 10^{-9} \mathrm{C} \end{aligned} $$

$$ \therefore \quad C_{p}=24 \times 10^{-12} \mathrm{~F} $$

[CBSE Marking Scheme 2017]

  1. Two identical parallel plate capacitors $A$ and $B$ are connected to a battery of $V$ volts with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $\kappa$. Find the ratio of the total electrostatic energy stored in bothcapacitors before and after the introduction of the dielectric.

[Topper’s Answers 2017]

[AI Q. 14. Two parallel plate capacitors $X$ and $Y$ have the same area of plates and same separation between them. $X$ has air between the plates while $Y$ contains a dielectric medium of $\varepsilon_{r}=4$.

(i) Calculate capacitance of each capaciton i equivalent capacitance of the combination is $4 \mu \mathrm{F}$.

(ii) Calculate the potential difference between the plates of $X$ and $Y$.

(iii) Estimate the ratio of electrostaticengy stored in $X$ and $Y$.

U] [Delhi I, II, III 2016, Set D, 2009, Set F 2008]

Ans. (i) Let

$$ \begin{aligned} & C_{X}=C \ & C_{Y}=4 C \end{aligned} $$

(as it has a dielectricmedium of $\varepsilon_{r}=4$ )

For series combination of two capacitors

$$ \frac{1}{C}=\frac{1}{C_{X}}+\frac{1}{C_{Y}} $$

$\Rightarrow \quad \frac{1}{4 \mu \mathrm{F}}=\frac{1}{\mathrm{C}}+\frac{1}{4 \mathrm{C}}$

$$ \frac{1}{4 \mu \mathrm{F}}=\frac{5}{4 \mathrm{C}} $$

$\Rightarrow \quad C=5 \mu \mathrm{F}$

Hence $\quad C_{X}=5 \mu \mathrm{F}$

(ii) Total charge

$$ C_{Y}=20 \mu \mathrm{F} $$

$$ \begin{aligned} & =4 \mu \mathrm{F} \times 15 \mathrm{~V}=60 \mu \mathrm{C} \ V_{X} & =\frac{Q}{C_{X}}=\frac{60 \mu \mathrm{C}}{5 \mu \mathrm{F}}=12 \mathrm{~V} \end{aligned} $$

$$ V_{Y}=\frac{Q}{C_{Y}}=\frac{60 \mu \mathrm{C}}{20 \mu \mathrm{F}}=3 \mathrm{~V} $$

(iii)

$$ =\frac{\overline{2 C_{X}}}{\frac{Q^{2}}{2 C_{x}}}=\frac{C_{Y}}{C_{X}}=\frac{20}{5}=4: 1 \quad \mathbf{1} $$

(Also accept any other correct alternative method)

[CBSE Marking Scheme 2016]

. A parallel plate capacitor, of capacitance $20 \mu \mathrm{F}$, is connected to a $100 \mathrm{~V}$, supply. After sometime, the battery is disconnected, and the space, between the plates of the capacitor is filled with a dielectric of dielectric constant 5 . Calculate the energy stored in the capacitor.

(i) Before

(ii) After the dielectric has been put in between its plates.

A [O.D. Comptt. I, II, III, 2016]

Ans. Charge stored, $Q=C V=20 \times 10^{-6} \times 100 \mathrm{C} \quad 1 / 2$

$$ =2000 \mu \mathrm{C} $$

New value of capacitance

$$ \begin{aligned} & =5 \times 20 \mu \mathrm{F} \ & =100 \mu \mathrm{F} \end{aligned} $$

Energy stored in a capacitor

$$ =\frac{1}{2} \frac{Q^{2}}{C}\left(=\frac{1}{2} C V^{2}=\frac{1}{2} Q V\right) \quad 1 / 2 $$

(i) Energy stored before dielectric

$$ \begin{align*} & =\frac{1}{2} \times \frac{\left[2000 \times 10^{-6}\right] \times\left(2000 \times 10^{-6}\right)}{20 \times 10^{-6}} \ & =0.1 \mathrm{~J} \tag{1} \end{align*} $$

(ii) Energy stored after the dielectric is introduced ( $\because$ there is no change in the value of $Q$ )

$$ \begin{align*} & =\frac{1}{2} \times \frac{2000 \times 10^{-6} \times 2000 \times 10^{-6}}{100 \times 10^{-6}} 1 / 2 \ & =0.02 \mathrm{~J} \ & 1 / 2 \end{align*} $$

[CBSE Marking Scheme, 2016]



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