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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Seven capacitors, each of capacitance 2 $$\mu $$F, are to be connected in a configuration to obtain an effective capacitance of $$\left( {{6 \over {13}}} \right)\mu F.$$ Which of the combinations, shown in figures below, will achieve the desired value

A

B

C

D

C_{eq} = $${6 \over {13}}$$$$\mu $$F

Therefore three capacitors most be in parallel to get 6 in

$${1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C}$$

C_{eq} = $${{3C} \over {13}}$$ = $${6 \over {13}}$$$$\mu $$F

Therefore three capacitors most be in parallel to get 6 in

$${1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C}$$

C

2

MCQ (Single Correct Answer)

In the figure shown below, the charge on the left plate of the 10$$\mu $$F capacitor is –30$$\mu $$C. The charge on the right plate of the 6 $$\mu $$F capacitor is :

A

+ 12 $$\mu $$C

B

+ 18 $$\mu $$C

C

$$-$$ 18 $$\mu $$C

D

$$-$$ 12 $$\mu $$C

6$$\mu $$F & 4$$\mu $$F are in parallel & total charge on this combination is 30 $$\mu $$C

$$ \therefore $$ Charge on 6$$\mu $$F capacitor = $${6 \over {6 + 4}} \times 30$$

= 18 $$\mu $$C

Since charge is asked on right plate therefore is +18$$\mu $$C

3

MCQ (Single Correct Answer)

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

A

508 pJ

B

692 pJ

C

560 pJ

D

600 pJ

Initial energy of capacitor

U_{i} = $${1 \over 2}$$ $${{{v^2}} \over c}$$

= $${1 \over 2}$$ $$ \times $$ $${{120 \times 120} \over {12}}$$ = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

U_{f} = $${1 \over 2}{{{v^2}} \over c}$$

= $${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$$ = 92

W + U_{f} = U_{i}

W = 508 J

U

= $${1 \over 2}$$ $$ \times $$ $${{120 \times 120} \over {12}}$$ = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

U

= $${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$$ = 92

W + U

W = 508 J

4

MCQ (Single Correct Answer)

A parallel plate capacitor is of area 6 cm^{2}
and a separation 3 mm. The gap is filled with three dielectric
materials of equal thickness (see figure) with dielectric constants K_{1} = 10, K_{2} = 12 and K_{3} = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be -

A

12

B

36

C

14

D

4

Let dielectric constant of material used be K.

$$ \therefore $$ $${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$$

$$ \Rightarrow $$ K $$=$$ 12

$$ \therefore $$ $${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$$

$$ \Rightarrow $$ K $$=$$ 12

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