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*Answer can only contain numeric values

QUESTION: 1

A leaky parallel plate capacitor is filled completely with a material having dielectric constant **K****= 5** and electrical conductivity **σ = 7.4 × 10 ^{–}^{12}Ω^{–1}m^{–1}**. If the charge on the capacitor at the instant

Solution:

Now current as function of time

or

The correct answer is: 0.193

*Answer can only contain numeric values

QUESTION: 2

In the figure shown find the equivalent capacitance between terminals * A* and

Solution:

The correct answer is: 13

*Answer can only contain numeric values

QUESTION: 3

A capacitor of capacitance 5µ*F* is connected to a source of constant emf of 200 *V*. Then the switch was shifted to contact 2 from contact 1. Find the amount of heat generated (in ** mJ**) in the 400Ω resistance.

Solution:

Potential energy stored in the capacitor,

During discharging this 0.1J, will distribute in direct ratio of resistance

The correct answer is: 44.4

*Answer can only contain numeric values

QUESTION: 4

A 8µ*F* capacitor **C**_{1} is charged to ** V_{0}** = 120

What is the stored energy (in units of 10

Solution:

The correct answer is: 3.84

*Answer can only contain numeric values

QUESTION: 5

In the circuit shown, a time varying voltage * V* = 2000

Solution:

At * t* = 5 ms,

Further ** q = CV **= (300 × 10

The correct answer is: 0.6

*Answer can only contain numeric values

QUESTION: 6

A parallel plate capacitor is to be designed which is to be connected across 1*kV* potential difference. The dielectric material which is to be filled between the plates has dielectric constant K = 6π and dielectric strength 10^{7}*V*/*m*. For safely the electric field is never to exceed 10% of the dielectric strength. With such specifications, if we want a capacitor of capacitance 50*pF*, what minimum area (in *mm*^{2}) of plates is required for safe working?

Solution:

The correct answer is: 300

*Answer can only contain numeric values

QUESTION: 7

In the figure shown the capacitor is initially uncharged. The current in (** R_{3} = R**) at time

Solution:

Applying Kirchoff’s law in

Loop1

Loop 2

eliminating *i*_{1} from (1) and (2)

*A* + *B* = 2 + 3 = 5

The correct answer is: 5

*Answer can only contain numeric values

QUESTION: 8

In the circuit shown a charged capacitor ** C_{1}** = 3

Solution:

The charge on the capacitor when current reaches *I*_{0}

*q*_{0} = (*I*_{0}*R*)·*C*_{1}

When the switch is in position 2, this charge is shared with capacitor ** C_{2}** and at steady state potential across

The correct answer is: 4

*Answer can only contain numeric values

QUESTION: 9

In the circuit shown the capacitors are initially uncharged. In a certain time the capacitor of capacitance 2µ*F* gets a charge 20*µC*. In that time interval, find the heat produced (in * µJ*) in each resistor 6

Solution:

Energy taken from cell = 20 × 30 µ*J*

= 600 µ*J*

Energy stored in capacitors

∴ Heat produced in resistors = 600 – 150 = 450 µ*J*

Divide this heat in **2**** Ω** and (equivalent of

which is 225µ

∴ Heat produced in

Further divide 225 µ

Heat (in **2*** Ω* ) 225µ

The correct answer is: 75

*Answer can only contain numeric values

QUESTION: 10

Given that ** C_{A}** = 1µ

Solution:

–*q*_{1} – *q*_{3} = 10 – 120 = – 110 ...(1)

–*q*_{2} + *q*_{3} = – 80 + 120 = + 40 ...(2)

In the final state

Solving we get q_{3} = 65μC

The charge on lower plate of capacitor *C _{C}* changes from –120µ

Hence the charge flowing through shown connecting wire is

(120 – 65) = 55 µ

final charges

Heat produced =

= 3025 µ*J*

The correct answer is: 75

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