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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359303 Assertion :
If three capacitors of capacitance $C_{1} < C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $C_{p} > C_{3}$
Reason :
$\frac{1}{C_{P}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For parallel connection
$C_{p} = C_{1} + C_{2} + C_{3} \Rightarrow C_{p} > C_{3}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359304 Four metal plates are arranged as shown. Capacitance between $X$ and $Y$ ( $A$ $\to$ Area each plate, d $\to$ distance between the plates)
supporting img

1

\frac{3}{2} \frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{2}{3} \frac{ε_{0} A}{d}

4

\frac{3 ε_{0} A}{d}

Explanation:

Given capacitors can be rearranged as
supporting img
Here $C = \frac{ε_{0} A}{d} ∴ C_{X Y} = \frac{2}{3} C = \frac{2}{3} \frac{ε_{0} A}{d}$

KCET - 2016

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359305 The ratio of equivalent capacitance between points $P$ and $Q$ in the following cases is
supporting img

1

1 : 2 : 3

2

3 : 2 : 1

3

1 : 1 : 1

4

1 : 1 : 2

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359306 The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances $1 p F, 2 p F a n d 4 p F$ is

1

\frac{7}{4} p F

2

\frac{4}{7} p F

3

1 p F

4

2 p F

Explanation:

$C_{1} = 1 p F, C_{2} = 2 p F a n d C_{3} = 4 p F$
Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance $C$ ' will be,
$\frac{1}{C^{'}} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} o r C^{'} = \frac{4}{7} p F$

KCET - 2017

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359303 Assertion :
If three capacitors of capacitance $C_{1} < C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $C_{p} > C_{3}$
Reason :
$\frac{1}{C_{P}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For parallel connection
$C_{p} = C_{1} + C_{2} + C_{3} \Rightarrow C_{p} > C_{3}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359304 Four metal plates are arranged as shown. Capacitance between $X$ and $Y$ ( $A$ $\to$ Area each plate, d $\to$ distance between the plates)
supporting img

1

\frac{3}{2} \frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{2}{3} \frac{ε_{0} A}{d}

4

\frac{3 ε_{0} A}{d}

Explanation:

Given capacitors can be rearranged as
supporting img
Here $C = \frac{ε_{0} A}{d} ∴ C_{X Y} = \frac{2}{3} C = \frac{2}{3} \frac{ε_{0} A}{d}$

KCET - 2016

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359305 The ratio of equivalent capacitance between points $P$ and $Q$ in the following cases is
supporting img

1

1 : 2 : 3

2

3 : 2 : 1

3

1 : 1 : 1

4

1 : 1 : 2

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359306 The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances $1 p F, 2 p F a n d 4 p F$ is

1

\frac{7}{4} p F

2

\frac{4}{7} p F

3

1 p F

4

2 p F

Explanation:

$C_{1} = 1 p F, C_{2} = 2 p F a n d C_{3} = 4 p F$
Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance $C$ ' will be,
$\frac{1}{C^{'}} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} o r C^{'} = \frac{4}{7} p F$

KCET - 2017

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359303 Assertion :
If three capacitors of capacitance $C_{1} < C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $C_{p} > C_{3}$
Reason :
$\frac{1}{C_{P}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For parallel connection
$C_{p} = C_{1} + C_{2} + C_{3} \Rightarrow C_{p} > C_{3}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359304 Four metal plates are arranged as shown. Capacitance between $X$ and $Y$ ( $A$ $\to$ Area each plate, d $\to$ distance between the plates)
supporting img

1

\frac{3}{2} \frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{2}{3} \frac{ε_{0} A}{d}

4

\frac{3 ε_{0} A}{d}

Explanation:

Given capacitors can be rearranged as
supporting img
Here $C = \frac{ε_{0} A}{d} ∴ C_{X Y} = \frac{2}{3} C = \frac{2}{3} \frac{ε_{0} A}{d}$

KCET - 2016

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359305 The ratio of equivalent capacitance between points $P$ and $Q$ in the following cases is
supporting img

1

1 : 2 : 3

2

3 : 2 : 1

3

1 : 1 : 1

4

1 : 1 : 2

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359306 The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances $1 p F, 2 p F a n d 4 p F$ is

1

\frac{7}{4} p F

2

\frac{4}{7} p F

3

1 p F

4

2 p F

Explanation:

$C_{1} = 1 p F, C_{2} = 2 p F a n d C_{3} = 4 p F$
Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance $C$ ' will be,
$\frac{1}{C^{'}} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} o r C^{'} = \frac{4}{7} p F$

KCET - 2017

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359303 Assertion :
If three capacitors of capacitance $C_{1} < C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $C_{p} > C_{3}$
Reason :
$\frac{1}{C_{P}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

For parallel connection
$C_{p} = C_{1} + C_{2} + C_{3} \Rightarrow C_{p} > C_{3}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359304 Four metal plates are arranged as shown. Capacitance between $X$ and $Y$ ( $A$ $\to$ Area each plate, d $\to$ distance between the plates)
supporting img

1

\frac{3}{2} \frac{ε_{0} A}{d}

2

\frac{2 ε_{0} A}{d}

3

\frac{2}{3} \frac{ε_{0} A}{d}

4

\frac{3 ε_{0} A}{d}

Explanation:

Given capacitors can be rearranged as
supporting img
Here $C = \frac{ε_{0} A}{d} ∴ C_{X Y} = \frac{2}{3} C = \frac{2}{3} \frac{ε_{0} A}{d}$

KCET - 2016

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359305 The ratio of equivalent capacitance between points $P$ and $Q$ in the following cases is
supporting img

1

1 : 2 : 3

2

3 : 2 : 1

3

1 : 1 : 1

4

1 : 1 : 2

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359306 The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances $1 p F, 2 p F a n d 4 p F$ is

1

\frac{7}{4} p F

2

\frac{4}{7} p F

3

1 p F

4

2 p F

Explanation:

$C_{1} = 1 p F, C_{2} = 2 p F a n d C_{3} = 4 p F$
Minimum effective capacitance can be obtained when capacitors are connected in series combination. The effective capacitance $C$ ' will be,
$\frac{1}{C^{'}} = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} o r C^{'} = \frac{4}{7} p F$

KCET - 2017

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