## In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations…

Q: In the given figure the capacitor of plate area A is charged upto charge q. The ratio of elongations (neglect force of gravity) in springs C and D at equilibrium position is

(a) $\displaystyle \frac{k_1}{k_2}$

(b) $\displaystyle \frac{k_2}{k_1}$

(c) $\displaystyle k_1 k_2$

(d) $\displaystyle \sqrt{ \frac{k_1}{k_2}}$

Ans: (b)
Sol: Force acting between plates of capacitor is attractive & it is

$\displaystyle F = \frac{q^2}{2 \epsilon_0 A}$

$\displaystyle k_1 x_1 = k_2 x_2$

$\displaystyle \frac{x_1}{x_2} = \frac{k_2}{k_1}$

## In figure, given C1 = 3 μF, C2 = 5 μF, C3 = 9 μF, and C4 = 13 μF . What is the potential difference…

Q: In figure, given C1 = 3 μF, C2 = 5 μF, C3 = 9 μF, and C4 = 13 μF . What is the potential difference between points A and B ?

(a) 13 V

(b) 9 V

(c) 0 V

(d) 11 V

Ans: (a)
Sol: V1 + V2 = 22 …(i)

9 V1 = 13 V2 …(ii)

## Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between …

Q: Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be (charge on each condenser is q0; k = dielectric constant)

(a) $\displaystyle \frac{2 q_0}{1+\frac{1}{k}}$

(b) $\displaystyle \frac{ q_0}{1+\frac{1}{k}}$

(c) $\displaystyle \frac{2 q_0}{1+k}$

(d) $\displaystyle \frac{ q_0}{1+k}$

Ans: (a)
Sol: $q_0/C + q_0/C = 2$ …(i)

After introduction of slab , let charge be q1

$q_1/C + q_1/kC = 2$ …(ii)

## The equivalent capacitance CAB of the circuit shown in the figure is…

Q: The equivalent capacitance CAB of the circuit shown in the figure is

(a) 5/4 C

(b) 4/5 C

(c) 2C

(d) C

Ans:(a)

## In the given network of capacitors as shown in figure, given that C1 = C2 = C3 = 4000 pF and …

Q: In the given network of capacitors as shown in figure, given that C1 = C2 = C3 = 4000 pF and C4= C5 = C6 = 200 pF. The effective capacitance of the circuit between X and Y is

(a) 810 pF

(b) 205 pF

(c) 600 pF

(d) 410 pF

Ans: (d)