16. ⇒ (MHT CET 2023 9th May Morning Shift )

The ratio of potential difference that must be applied across parallel and series combination of two capacitors $C_{1}$ and $C_{2}$ with their capacitance in the ratio $1 : 2$ so that energy stored in these two cases becomes same is

A. $3 : \sqrt{2}$

B. $\sqrt{2} : 3$

C. $2 : 9$

D. $9 : 2$

Correct Option is (B)

Given: $C_{1} : C_{2} = 1 : 2$

$\begin{aligned} ∴ C_{2} & = 2 C_{1} \\ C_{P} & = C_{1} + C_{2} = C_{1} + 2 C_{1} = 3 C_{1} \\ C_{S} & = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{2 C_{1}^{2}}{3 C_{1}} = \frac{2}{3} C_{1} \end{aligned}$

Let $V_{P}$ and $V_{S}$ be the potentials applied across the parallel and series combinations respectively, then using $E = \frac{1}{2} {CV}^{2}$ we can write,

$\begin{array}{ll} \frac{1}{2} C_{P} V_{P}^{2} = \frac{1}{2} C_{S} V_{S}^{2} \\ ∴ & \frac{V_{P}^{2}}{V_{S}^{2}} = \frac{C_{S}}{C_{P}} \\ \frac{V_{P}}{V_{S}} = \sqrt{\frac{\frac{2}{3} C_{1}}{3 C_{1}}} = \frac{\sqrt{2}}{3} \end{array}$

17. ⇒ (MHT CET 2023 9th May Morning Shift )

The potential energy of charged parallel plate capacitor is $v_{0}$ . If a slab of dielectric constant $K$ is inserted between the plates, then the new potential energy will be

A. $\frac{v_{0}}{K}$

B. $v_{0} K^{2}$

C. $\frac{v_{0}}{K^{2}}$

D. $v_{0}^{2}$

Correct Option is (A)

We know, $v_{0} = \frac{Q^{2}}{2 C}$

On inserting the slab of dielectric constant $k$ , the new capacitance $C^{'} = KC$

$∴$ New potential energy $v_{0}^{'} = \frac{Q^{2}}{2 C^{'}}$

$v_{0}^{1} = \frac{Q^{2}}{2 KC} = \frac{v_{0}}{K}$

18. ⇒ (MHT CET 2021 21th September Evening Shift )

Two identical parallel plate air capacitors are connected in series to a battery of emf ' $V$ '. If one of the capacitor is inserted in liquid of dielectric constant ' $K$ ' then, potential difference of the other capacitor will become

A. $\frac{K - 1}{KV}$

B. $\frac{K + 1}{KV}$

C. $(\frac{KV}{K + 1})$

D. $\frac{KV}{K - 1}$

Correct Option is (C)

Let the capacitance of capacitors are $C$ initially.

So, afterward $C_{1} = kC$ and $C_{2} = C$

p.d. across capacitor $C_{2}$ is

$V_{2} = \frac{C_{2} V}{C_{1} + C_{2}} = \frac{k C V}{C + k C} = \frac{k V}{k + 1}$

19. ⇒ (MHT CET 2021 21th September Evening Shift )

A condenser of capacity ' $C_{1}$ ' is charged to potential ' $V_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $C_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is

A. $\frac{V_{1} C_{2}}{C_{1}}$

B. $\frac{C_{2}}{C_{1} + C_{2}}$

C. $\frac{C_{1} V_{1}}{C_{2}}$

D. $\frac{C_{1} V_{1}}{C_{1} + C_{2}}$

Correct Option is (D)

The charge stored by the charged condenser is

$Q = C_{1} V_{1}$

When the uncharged condenser is connected in parallel, the effective capacitance becomes $(C_{1} + C_{2})$

Hence the potential $V_{2} = \frac{Q}{C_{1} + C_{2}} = \frac{C_{1} V_{1}}{C_{1} + C_{2}}$

20. ⇒ (MHT CET 2021 21th September Morning Shift )

The resultant capacity between points A and B in the given circuit is

MHT CET 2021 21th September Morning Shift Physics - Capacitor Question 20 English

A. C

B. $\frac{C}{3}$

C. 3C

D. 2C

Correct Option is (C)

MHT CET 2021 21th September Morning Shift Physics - Capacitor Question 20 English Explanation

$C_{1}$ and $C_{2}$ are in parallel

Their equivalent capacitance $C_{7} = 2 C$

$C_{7}$ and $C_{3}$ are in series

Their equivlanet capacitance is $C_{8} = C$

$C_{8}$ and $C_{4}$ are in parallel

Their equivalent capacitance is $C_{9} = 2 C$

$C_{9}$ and $C_{5}$ are in series. Their equivalent capacitance $C_{10} = C$

$C_{10}$ and $C_{6}$ are in parallel. Their equivalent capacitance is $3 C$ .

Hence equivalent capacitance between $A$ and $B$ is $3 C$ .

⇒Electrostatics