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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359311 Find equivalent Capacitance between point $A$ and $B$ if Capacitance between any two plates is $C$ .
supporting img

1

\frac{C}{n + 1}

2

\frac{C}{2 n + 1}

3

\frac{C}{2 n - 1}

4

\frac{C}{n - 1}

Explanation:

There are total $(n - 1)$ capacitors which are in series.
So $\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \dots (n - 1)$ times
$\begin{aligned} \frac{1}{C_{e q}} = \frac{(n - 1)}{C} \\ \Rightarrow C_{e q} = \frac{C}{n - 1} \end{aligned}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359312 The ratio of the resultant capacities when three capacitors of $2 μ F, 4 μ F, a n d 6 μ F$ are connected first in series and then in parallel is

1

1 : 11

2

11 : 1

3

12 : 1

4

1 : 12

Explanation:

$\frac{1}{C_{1}} = \frac{1}{2 μ F} + \frac{1}{4 μ F} + \frac{1}{6 μ F},$
$C_{2} = 2 μ F + 4 μ F + 6 μ F$
$C_{1} : C_{2} = 1 : 11$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359313 The following arrangement consists of four plates. The area of each plate is $A$ and separation between successive plates is $d$ . The ratio of effective capacitance between $P$ and $Q$ as shown in figure (i) and (ii) is
supporting img

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{4}{3}

4

1

Explanation:

For figure (i)
The arrangement will work as a system of three capacitors connected in parallel as shown in the figure.
The capacitance of each capactior, $C = \frac{ε_{0} A}{d}$
The effective capacitance between P and Q is
$C_{1} = 2 \frac{ε_{0} A}{d}$
For figure (ii)
The arrangement will work as two capacitors connected in parallel as shown in figure.
The effective capacitance between P and Q is
$C_{2} = \frac{3 ε_{0} A}{d} ∴ \frac{C_{1}}{C_{2}} = \frac{2}{3}$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359314 Three uncharged capacitors of capacitances $C_{1}, C_{2} a n d C_{3}$ are connected as shown in the figure. The potential at $O$ will be:
supporting img

1

\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

2

\frac{V_{1} C_{1}^{2} + V_{2} C_{2}^{2} + V_{3} C_{3}^{2}}{C_{1}^{2} + C_{2}^{2} + C_{3}^{2}}

3

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

4

\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

Explanation:

Let $V$ be the potential at the junction $O$ . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge $C_{1}$ ) + (Charge on $C_{2}$ ) = (Charge on $C_{3}$ )
$q_{1} + q_{2} = (q_{1} + q_{2})$
$C_{1} (V - V_{1}) + C_{2} (V - V_{2}) = C_{3} (V_{3} - V)$
supporting img
$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{2}} = \frac{Σ V_{i} C_{i}}{Σ C_{i}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359315 In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
supporting img

1

2 μ F

2

1 μ F

3

0.5 μ F

4

4 μ F

Explanation:

supporting img
Given circuit is balanced Wheatstone bridge so the middle capacitor can be disconnected.

$C_{A B} = 1 + 1 = 2 μ F$

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359311 Find equivalent Capacitance between point $A$ and $B$ if Capacitance between any two plates is $C$ .
supporting img

1

\frac{C}{n + 1}

2

\frac{C}{2 n + 1}

3

\frac{C}{2 n - 1}

4

\frac{C}{n - 1}

Explanation:

There are total $(n - 1)$ capacitors which are in series.
So $\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \dots (n - 1)$ times
$\begin{aligned} \frac{1}{C_{e q}} = \frac{(n - 1)}{C} \\ \Rightarrow C_{e q} = \frac{C}{n - 1} \end{aligned}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359312 The ratio of the resultant capacities when three capacitors of $2 μ F, 4 μ F, a n d 6 μ F$ are connected first in series and then in parallel is

1

1 : 11

2

11 : 1

3

12 : 1

4

1 : 12

Explanation:

$\frac{1}{C_{1}} = \frac{1}{2 μ F} + \frac{1}{4 μ F} + \frac{1}{6 μ F},$
$C_{2} = 2 μ F + 4 μ F + 6 μ F$
$C_{1} : C_{2} = 1 : 11$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359313 The following arrangement consists of four plates. The area of each plate is $A$ and separation between successive plates is $d$ . The ratio of effective capacitance between $P$ and $Q$ as shown in figure (i) and (ii) is
supporting img

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{4}{3}

4

1

Explanation:

For figure (i)
The arrangement will work as a system of three capacitors connected in parallel as shown in the figure.
The capacitance of each capactior, $C = \frac{ε_{0} A}{d}$
The effective capacitance between P and Q is
$C_{1} = 2 \frac{ε_{0} A}{d}$
For figure (ii)
The arrangement will work as two capacitors connected in parallel as shown in figure.
The effective capacitance between P and Q is
$C_{2} = \frac{3 ε_{0} A}{d} ∴ \frac{C_{1}}{C_{2}} = \frac{2}{3}$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359314 Three uncharged capacitors of capacitances $C_{1}, C_{2} a n d C_{3}$ are connected as shown in the figure. The potential at $O$ will be:
supporting img

1

\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

2

\frac{V_{1} C_{1}^{2} + V_{2} C_{2}^{2} + V_{3} C_{3}^{2}}{C_{1}^{2} + C_{2}^{2} + C_{3}^{2}}

3

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

4

\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

Explanation:

Let $V$ be the potential at the junction $O$ . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge $C_{1}$ ) + (Charge on $C_{2}$ ) = (Charge on $C_{3}$ )
$q_{1} + q_{2} = (q_{1} + q_{2})$
$C_{1} (V - V_{1}) + C_{2} (V - V_{2}) = C_{3} (V_{3} - V)$
supporting img
$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{2}} = \frac{Σ V_{i} C_{i}}{Σ C_{i}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359315 In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
supporting img

1

2 μ F

2

1 μ F

3

0.5 μ F

4

4 μ F

Explanation:

supporting img
Given circuit is balanced Wheatstone bridge so the middle capacitor can be disconnected.

$C_{A B} = 1 + 1 = 2 μ F$

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359311 Find equivalent Capacitance between point $A$ and $B$ if Capacitance between any two plates is $C$ .
supporting img

1

\frac{C}{n + 1}

2

\frac{C}{2 n + 1}

3

\frac{C}{2 n - 1}

4

\frac{C}{n - 1}

Explanation:

There are total $(n - 1)$ capacitors which are in series.
So $\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \dots (n - 1)$ times
$\begin{aligned} \frac{1}{C_{e q}} = \frac{(n - 1)}{C} \\ \Rightarrow C_{e q} = \frac{C}{n - 1} \end{aligned}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359312 The ratio of the resultant capacities when three capacitors of $2 μ F, 4 μ F, a n d 6 μ F$ are connected first in series and then in parallel is

1

1 : 11

2

11 : 1

3

12 : 1

4

1 : 12

Explanation:

$\frac{1}{C_{1}} = \frac{1}{2 μ F} + \frac{1}{4 μ F} + \frac{1}{6 μ F},$
$C_{2} = 2 μ F + 4 μ F + 6 μ F$
$C_{1} : C_{2} = 1 : 11$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359313 The following arrangement consists of four plates. The area of each plate is $A$ and separation between successive plates is $d$ . The ratio of effective capacitance between $P$ and $Q$ as shown in figure (i) and (ii) is
supporting img

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{4}{3}

4

1

Explanation:

For figure (i)
The arrangement will work as a system of three capacitors connected in parallel as shown in the figure.
The capacitance of each capactior, $C = \frac{ε_{0} A}{d}$
The effective capacitance between P and Q is
$C_{1} = 2 \frac{ε_{0} A}{d}$
For figure (ii)
The arrangement will work as two capacitors connected in parallel as shown in figure.
The effective capacitance between P and Q is
$C_{2} = \frac{3 ε_{0} A}{d} ∴ \frac{C_{1}}{C_{2}} = \frac{2}{3}$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359314 Three uncharged capacitors of capacitances $C_{1}, C_{2} a n d C_{3}$ are connected as shown in the figure. The potential at $O$ will be:
supporting img

1

\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

2

\frac{V_{1} C_{1}^{2} + V_{2} C_{2}^{2} + V_{3} C_{3}^{2}}{C_{1}^{2} + C_{2}^{2} + C_{3}^{2}}

3

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

4

\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

Explanation:

Let $V$ be the potential at the junction $O$ . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge $C_{1}$ ) + (Charge on $C_{2}$ ) = (Charge on $C_{3}$ )
$q_{1} + q_{2} = (q_{1} + q_{2})$
$C_{1} (V - V_{1}) + C_{2} (V - V_{2}) = C_{3} (V_{3} - V)$
supporting img
$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{2}} = \frac{Σ V_{i} C_{i}}{Σ C_{i}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359315 In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
supporting img

1

2 μ F

2

1 μ F

3

0.5 μ F

4

4 μ F

Explanation:

supporting img
Given circuit is balanced Wheatstone bridge so the middle capacitor can be disconnected.

$C_{A B} = 1 + 1 = 2 μ F$

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359311 Find equivalent Capacitance between point $A$ and $B$ if Capacitance between any two plates is $C$ .
supporting img

1

\frac{C}{n + 1}

2

\frac{C}{2 n + 1}

3

\frac{C}{2 n - 1}

4

\frac{C}{n - 1}

Explanation:

There are total $(n - 1)$ capacitors which are in series.
So $\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \dots (n - 1)$ times
$\begin{aligned} \frac{1}{C_{e q}} = \frac{(n - 1)}{C} \\ \Rightarrow C_{e q} = \frac{C}{n - 1} \end{aligned}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359312 The ratio of the resultant capacities when three capacitors of $2 μ F, 4 μ F, a n d 6 μ F$ are connected first in series and then in parallel is

1

1 : 11

2

11 : 1

3

12 : 1

4

1 : 12

Explanation:

$\frac{1}{C_{1}} = \frac{1}{2 μ F} + \frac{1}{4 μ F} + \frac{1}{6 μ F},$
$C_{2} = 2 μ F + 4 μ F + 6 μ F$
$C_{1} : C_{2} = 1 : 11$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359313 The following arrangement consists of four plates. The area of each plate is $A$ and separation between successive plates is $d$ . The ratio of effective capacitance between $P$ and $Q$ as shown in figure (i) and (ii) is
supporting img

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{4}{3}

4

1

Explanation:

For figure (i)
The arrangement will work as a system of three capacitors connected in parallel as shown in the figure.
The capacitance of each capactior, $C = \frac{ε_{0} A}{d}$
The effective capacitance between P and Q is
$C_{1} = 2 \frac{ε_{0} A}{d}$
For figure (ii)
The arrangement will work as two capacitors connected in parallel as shown in figure.
The effective capacitance between P and Q is
$C_{2} = \frac{3 ε_{0} A}{d} ∴ \frac{C_{1}}{C_{2}} = \frac{2}{3}$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359314 Three uncharged capacitors of capacitances $C_{1}, C_{2} a n d C_{3}$ are connected as shown in the figure. The potential at $O$ will be:
supporting img

1

\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

2

\frac{V_{1} C_{1}^{2} + V_{2} C_{2}^{2} + V_{3} C_{3}^{2}}{C_{1}^{2} + C_{2}^{2} + C_{3}^{2}}

3

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

4

\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

Explanation:

Let $V$ be the potential at the junction $O$ . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge $C_{1}$ ) + (Charge on $C_{2}$ ) = (Charge on $C_{3}$ )
$q_{1} + q_{2} = (q_{1} + q_{2})$
$C_{1} (V - V_{1}) + C_{2} (V - V_{2}) = C_{3} (V_{3} - V)$
supporting img
$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{2}} = \frac{Σ V_{i} C_{i}}{Σ C_{i}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359315 In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
supporting img

1

2 μ F

2

1 μ F

3

0.5 μ F

4

4 μ F

Explanation:

supporting img
Given circuit is balanced Wheatstone bridge so the middle capacitor can be disconnected.

$C_{A B} = 1 + 1 = 2 μ F$

NEET - 2024

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359311 Find equivalent Capacitance between point $A$ and $B$ if Capacitance between any two plates is $C$ .
supporting img

1

\frac{C}{n + 1}

2

\frac{C}{2 n + 1}

3

\frac{C}{2 n - 1}

4

\frac{C}{n - 1}

Explanation:

There are total $(n - 1)$ capacitors which are in series.
So $\frac{1}{C_{e q}} = \frac{1}{C} + \frac{1}{C} + \dots (n - 1)$ times
$\begin{aligned} \frac{1}{C_{e q}} = \frac{(n - 1)}{C} \\ \Rightarrow C_{e q} = \frac{C}{n - 1} \end{aligned}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359312 The ratio of the resultant capacities when three capacitors of $2 μ F, 4 μ F, a n d 6 μ F$ are connected first in series and then in parallel is

1

1 : 11

2

11 : 1

3

12 : 1

4

1 : 12

Explanation:

$\frac{1}{C_{1}} = \frac{1}{2 μ F} + \frac{1}{4 μ F} + \frac{1}{6 μ F},$
$C_{2} = 2 μ F + 4 μ F + 6 μ F$
$C_{1} : C_{2} = 1 : 11$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359313 The following arrangement consists of four plates. The area of each plate is $A$ and separation between successive plates is $d$ . The ratio of effective capacitance between $P$ and $Q$ as shown in figure (i) and (ii) is
supporting img

1

\frac{2}{3}

2

\frac{3}{2}

3

\frac{4}{3}

4

1

Explanation:

For figure (i)
The arrangement will work as a system of three capacitors connected in parallel as shown in the figure.
The capacitance of each capactior, $C = \frac{ε_{0} A}{d}$
The effective capacitance between P and Q is
$C_{1} = 2 \frac{ε_{0} A}{d}$
For figure (ii)
The arrangement will work as two capacitors connected in parallel as shown in figure.
The effective capacitance between P and Q is
$C_{2} = \frac{3 ε_{0} A}{d} ∴ \frac{C_{1}}{C_{2}} = \frac{2}{3}$
supporting img

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359314 Three uncharged capacitors of capacitances $C_{1}, C_{2} a n d C_{3}$ are connected as shown in the figure. The potential at $O$ will be:
supporting img

1

\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

2

\frac{V_{1} C_{1}^{2} + V_{2} C_{2}^{2} + V_{3} C_{3}^{2}}{C_{1}^{2} + C_{2}^{2} + C_{3}^{2}}

3

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

4

\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} + C_{2} C_{3} + C_{3} C_{1}}

Explanation:

Let $V$ be the potential at the junction $O$ . There is an independent loop in the circuit. The total charge on the independent loop is zero i.e., (Charge $C_{1}$ ) + (Charge on $C_{2}$ ) = (Charge on $C_{3}$ )
$q_{1} + q_{2} = (q_{1} + q_{2})$
$C_{1} (V - V_{1}) + C_{2} (V - V_{2}) = C_{3} (V_{3} - V)$
supporting img
$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{2}} = \frac{Σ V_{i} C_{i}}{Σ C_{i}}$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359315 In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
supporting img

1

2 μ F

2

1 μ F

3

0.5 μ F

4

4 μ F

Explanation:

supporting img
Given circuit is balanced Wheatstone bridge so the middle capacitor can be disconnected.

$C_{A B} = 1 + 1 = 2 μ F$

NEET - 2024