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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359298 Three capacitors each of capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

1

3 C, V / 3

2

C / 3, 3 V

3

3 C, 3 V

4

C / 3, V / 3

Explanation:

$C_{e q} = \frac{C}{3} V_{M a x} = V + V + V = 3 V$

NCERT Exemplar

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359299 A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 $c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is
supporting img

1

1.06 p F

2

4 p F

3

6.36 p F

4

12.72 p F

Explanation:

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
$C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 5 \times 10^{- 4}}{0.885 \times 10^{- 2}} = 0.5 p F$
The total capacity of 8 capacitors $= 8 C$
$= 8 \times 0.5 = 4 p F$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359300 The equivalent capacitance between $A$ and $B$ in the circuit given below is:
supporting img

1

3.6 μ F

2

2.4 μ F

3

4.9 μ F

4

5.4 μ F

Explanation:

supporting img
By joining the points having identical potentials

$\frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{e q} = \frac{12}{5} = 2.4 μ F$

JEE - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359301 For circuit, the equivalent capacitance between $P$ and $Q$ is
supporting img

1

6 C

2

4 C

3

\frac{3 C}{2}

4

\frac{6 C}{11}

Explanation:

supporting img
$\frac{1}{C_{e q}} = \frac{1}{3 C} + \frac{1}{2 C} + \frac{1}{C}$
$= \frac{2 + 3 + 6}{6 C} = \frac{11}{6 C}$
$\Rightarrow C_{e q} = \frac{6 C}{11}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359302 The equivalent capacitance of the system shown in the following circuit is:
supporting img

1

3 μ F

2

6 μ F

3

9 μ F

4

2 μ F

Explanation:

The two capacitors on right side are in parallel connection equivalent to $(3 + 3) = 6 μ F$ . Then, that $6 μ F$ is in series with $3 μ F$ of left side.
supporting img
$\Rightarrow C_{A B} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}}$
$\Rightarrow C_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μ F$
Correct option is (4).

NEET - 2023

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359298 Three capacitors each of capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

1

3 C, V / 3

2

C / 3, 3 V

3

3 C, 3 V

4

C / 3, V / 3

Explanation:

$C_{e q} = \frac{C}{3} V_{M a x} = V + V + V = 3 V$

NCERT Exemplar

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359299 A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 $c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is
supporting img

1

1.06 p F

2

4 p F

3

6.36 p F

4

12.72 p F

Explanation:

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
$C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 5 \times 10^{- 4}}{0.885 \times 10^{- 2}} = 0.5 p F$
The total capacity of 8 capacitors $= 8 C$
$= 8 \times 0.5 = 4 p F$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359300 The equivalent capacitance between $A$ and $B$ in the circuit given below is:
supporting img

1

3.6 μ F

2

2.4 μ F

3

4.9 μ F

4

5.4 μ F

Explanation:

supporting img
By joining the points having identical potentials

$\frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{e q} = \frac{12}{5} = 2.4 μ F$

JEE - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359301 For circuit, the equivalent capacitance between $P$ and $Q$ is
supporting img

1

6 C

2

4 C

3

\frac{3 C}{2}

4

\frac{6 C}{11}

Explanation:

supporting img
$\frac{1}{C_{e q}} = \frac{1}{3 C} + \frac{1}{2 C} + \frac{1}{C}$
$= \frac{2 + 3 + 6}{6 C} = \frac{11}{6 C}$
$\Rightarrow C_{e q} = \frac{6 C}{11}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359302 The equivalent capacitance of the system shown in the following circuit is:
supporting img

1

3 μ F

2

6 μ F

3

9 μ F

4

2 μ F

Explanation:

The two capacitors on right side are in parallel connection equivalent to $(3 + 3) = 6 μ F$ . Then, that $6 μ F$ is in series with $3 μ F$ of left side.
supporting img
$\Rightarrow C_{A B} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}}$
$\Rightarrow C_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μ F$
Correct option is (4).

NEET - 2023

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359298 Three capacitors each of capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

1

3 C, V / 3

2

C / 3, 3 V

3

3 C, 3 V

4

C / 3, V / 3

Explanation:

$C_{e q} = \frac{C}{3} V_{M a x} = V + V + V = 3 V$

NCERT Exemplar

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359299 A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 $c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is
supporting img

1

1.06 p F

2

4 p F

3

6.36 p F

4

12.72 p F

Explanation:

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
$C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 5 \times 10^{- 4}}{0.885 \times 10^{- 2}} = 0.5 p F$
The total capacity of 8 capacitors $= 8 C$
$= 8 \times 0.5 = 4 p F$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359300 The equivalent capacitance between $A$ and $B$ in the circuit given below is:
supporting img

1

3.6 μ F

2

2.4 μ F

3

4.9 μ F

4

5.4 μ F

Explanation:

supporting img
By joining the points having identical potentials

$\frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{e q} = \frac{12}{5} = 2.4 μ F$

JEE - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359301 For circuit, the equivalent capacitance between $P$ and $Q$ is
supporting img

1

6 C

2

4 C

3

\frac{3 C}{2}

4

\frac{6 C}{11}

Explanation:

supporting img
$\frac{1}{C_{e q}} = \frac{1}{3 C} + \frac{1}{2 C} + \frac{1}{C}$
$= \frac{2 + 3 + 6}{6 C} = \frac{11}{6 C}$
$\Rightarrow C_{e q} = \frac{6 C}{11}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359302 The equivalent capacitance of the system shown in the following circuit is:
supporting img

1

3 μ F

2

6 μ F

3

9 μ F

4

2 μ F

Explanation:

The two capacitors on right side are in parallel connection equivalent to $(3 + 3) = 6 μ F$ . Then, that $6 μ F$ is in series with $3 μ F$ of left side.
supporting img
$\Rightarrow C_{A B} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}}$
$\Rightarrow C_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μ F$
Correct option is (4).

NEET - 2023

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359298 Three capacitors each of capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

1

3 C, V / 3

2

C / 3, 3 V

3

3 C, 3 V

4

C / 3, V / 3

Explanation:

$C_{e q} = \frac{C}{3} V_{M a x} = V + V + V = 3 V$

NCERT Exemplar

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359299 A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 $c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is
supporting img

1

1.06 p F

2

4 p F

3

6.36 p F

4

12.72 p F

Explanation:

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
$C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 5 \times 10^{- 4}}{0.885 \times 10^{- 2}} = 0.5 p F$
The total capacity of 8 capacitors $= 8 C$
$= 8 \times 0.5 = 4 p F$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359300 The equivalent capacitance between $A$ and $B$ in the circuit given below is:
supporting img

1

3.6 μ F

2

2.4 μ F

3

4.9 μ F

4

5.4 μ F

Explanation:

supporting img
By joining the points having identical potentials

$\frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{e q} = \frac{12}{5} = 2.4 μ F$

JEE - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359301 For circuit, the equivalent capacitance between $P$ and $Q$ is
supporting img

1

6 C

2

4 C

3

\frac{3 C}{2}

4

\frac{6 C}{11}

Explanation:

supporting img
$\frac{1}{C_{e q}} = \frac{1}{3 C} + \frac{1}{2 C} + \frac{1}{C}$
$= \frac{2 + 3 + 6}{6 C} = \frac{11}{6 C}$
$\Rightarrow C_{e q} = \frac{6 C}{11}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359302 The equivalent capacitance of the system shown in the following circuit is:
supporting img

1

3 μ F

2

6 μ F

3

9 μ F

4

2 μ F

Explanation:

The two capacitors on right side are in parallel connection equivalent to $(3 + 3) = 6 μ F$ . Then, that $6 μ F$ is in series with $3 μ F$ of left side.
supporting img
$\Rightarrow C_{A B} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}}$
$\Rightarrow C_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μ F$
Correct option is (4).

NEET - 2023

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359298 Three capacitors each of capacitance $C$ and breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

1

3 C, V / 3

2

C / 3, 3 V

3

3 C, 3 V

4

C / 3, V / 3

Explanation:

$C_{e q} = \frac{C}{3} V_{M a x} = V + V + V = 3 V$

NCERT Exemplar

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359299 A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 $c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is
supporting img

1

1.06 p F

2

4 p F

3

6.36 p F

4

12.72 p F

Explanation:

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
$C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 5 \times 10^{- 4}}{0.885 \times 10^{- 2}} = 0.5 p F$
The total capacity of 8 capacitors $= 8 C$
$= 8 \times 0.5 = 4 p F$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359300 The equivalent capacitance between $A$ and $B$ in the circuit given below is:
supporting img

1

3.6 μ F

2

2.4 μ F

3

4.9 μ F

4

5.4 μ F

Explanation:

supporting img
By joining the points having identical potentials

$\frac{1}{C_{e q}} = \frac{1}{6} + \frac{1}{12} + \frac{1}{6} = \frac{5}{12} \Rightarrow C_{e q} = \frac{12}{5} = 2.4 μ F$

JEE - 2018

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359301 For circuit, the equivalent capacitance between $P$ and $Q$ is
supporting img

1

6 C

2

4 C

3

\frac{3 C}{2}

4

\frac{6 C}{11}

Explanation:

supporting img
$\frac{1}{C_{e q}} = \frac{1}{3 C} + \frac{1}{2 C} + \frac{1}{C}$
$= \frac{2 + 3 + 6}{6 C} = \frac{11}{6 C}$
$\Rightarrow C_{e q} = \frac{6 C}{11}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359302 The equivalent capacitance of the system shown in the following circuit is:
supporting img

1

3 μ F

2

6 μ F

3

9 μ F

4

2 μ F

Explanation:

The two capacitors on right side are in parallel connection equivalent to $(3 + 3) = 6 μ F$ . Then, that $6 μ F$ is in series with $3 μ F$ of left side.
supporting img
$\Rightarrow C_{A B} = \frac{C_{1} \times C_{2}}{C_{1} + C_{2}}$
$\Rightarrow C_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 μ F$
Correct option is (4).

NEET - 2023