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Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $C_{1} / C_{2}$ is

1

\frac{1}{4}

2

\frac{1}{16}

3

\frac{1}{8}

4

\frac{1}{12}

Explanation:

: Let capacitance of each capacitor is c In series, net capacitance,
$C_{2} = \frac{C}{4}$
In Parallel, net capacitance
$C_{2} = 4 C$
The ratio of $\frac{C_{1}}{C_{2}}$ is,
$\frac{C_{1}}{C_{2}} = \frac{C}{4 \times 4 C} = \frac{1}{16}$
$\frac{C_{1}}{C_{2}} = \frac{1}{16}$

[WB JEE-2007]

Capacitance

165663 The effective capacitances of two capacitors are $3 μ F$ and $16 μ F$ , when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1

10 μ F, 6 μ F

2

8 μ F, 8 μ F

3

12 μ F, 4 μ F

4

1.2 μ F, 1.8 μ F

Explanation:

: Given,
$C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = 3 μ F$
$(C_{s}$ is equivalent capacitance of capacitor in series)
$C_{p} = C_{1} + C_{2} = 16 μ F$
$(C_{p}$ is equivalent capacitance of capacitor in Parallel)
From (i) \& (ii)
$C_{1} C_{2} = 3 \times 16 = 48$
From (ii) \& (iii)
From (iii)
$C_{1} + \frac{48}{C_{1}} = 16$
$C_{1}^{2} - 16 C_{1} + 48 = 0$
$C_{1}^{2} - 12 C_{1} - 4 C_{1} + 48 = 0$
$C_{1} (C_{1} - 12) - 4 (C_{1} - 12) = 0$
$(C_{1} - 12) (C_{1} - 4) = 0$
$C_{1} = 12 μ F, 4 μ F$
$12 \times C_{2} = 48$
$C_{2} = \frac{48}{12}$
$= 4 μ F$
$C_{1} = 12 μ F C_{2} = 4 μ F$

Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1

2 C

2

C / 2

3

4 C

4

C / 4

Explanation:

: Capacitance,
$C = \frac{ε_{0} A}{d}$
Area $(a) = A / 2$
Diameter $(d) = 2 d$
Then, $C^{'} = \frac{ε_{0} A}{2 \times 2 d}$
$C^{'} = \frac{ε_{0} A}{4 d} = \frac{C}{4}$
$C^{'} = \frac{C}{4}$

Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$ , where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $C_{0} =$ capacitance of capacitor with air as medium between plates.)

1

\frac{4 {KC}_{0}}{3 + K}

2

\frac{3 {KC}_{0}}{3 + K}

3

\frac{3 + K}{4 {KC}_{0}}

4

\frac{K}{4 + K}

Explanation:

: Given that,
Thickness of slab $(t) = \frac{3}{4} d$
The capacitance of a parallel plate capacitor in the absence of the dielectric is-
$C_{0} = \frac{ε_{0} A}{d}$
If a dielectric slab is partially filled between the plates of capacitor, then
$C = \frac{ε_{0} A}{[d - t + \frac{t}{K}]}$
$C = \frac{ε_{0} A}{(d - \frac{3}{4} d) + (\frac{3 d}{4 K})}$
$C = \frac{ε_{0} A}{\frac{d}{4} + \frac{3 d}{4 K}}$
$C = \frac{4 K ε_{0} A}{d (K + 3)}$
$C = \frac{4 K ε_{0} A}{d (K + 3)} \times \frac{d}{ε_{0} A}$
$C = \frac{4 KC C_{0}}{(3 + K)}$

[JEE Main-28.07.2022

Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $C_{1} / C_{2}$ is

1

\frac{1}{4}

2

\frac{1}{16}

3

\frac{1}{8}

4

\frac{1}{12}

Explanation:

: Let capacitance of each capacitor is c In series, net capacitance,
$C_{2} = \frac{C}{4}$
In Parallel, net capacitance
$C_{2} = 4 C$
The ratio of $\frac{C_{1}}{C_{2}}$ is,
$\frac{C_{1}}{C_{2}} = \frac{C}{4 \times 4 C} = \frac{1}{16}$
$\frac{C_{1}}{C_{2}} = \frac{1}{16}$

[WB JEE-2007]

Capacitance

165663 The effective capacitances of two capacitors are $3 μ F$ and $16 μ F$ , when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1

10 μ F, 6 μ F

2

8 μ F, 8 μ F

3

12 μ F, 4 μ F

4

1.2 μ F, 1.8 μ F

Explanation:

: Given,
$C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = 3 μ F$
$(C_{s}$ is equivalent capacitance of capacitor in series)
$C_{p} = C_{1} + C_{2} = 16 μ F$
$(C_{p}$ is equivalent capacitance of capacitor in Parallel)
From (i) \& (ii)
$C_{1} C_{2} = 3 \times 16 = 48$
From (ii) \& (iii)
From (iii)
$C_{1} + \frac{48}{C_{1}} = 16$
$C_{1}^{2} - 16 C_{1} + 48 = 0$
$C_{1}^{2} - 12 C_{1} - 4 C_{1} + 48 = 0$
$C_{1} (C_{1} - 12) - 4 (C_{1} - 12) = 0$
$(C_{1} - 12) (C_{1} - 4) = 0$
$C_{1} = 12 μ F, 4 μ F$
$12 \times C_{2} = 48$
$C_{2} = \frac{48}{12}$
$= 4 μ F$
$C_{1} = 12 μ F C_{2} = 4 μ F$

Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1

2 C

2

C / 2

3

4 C

4

C / 4

Explanation:

: Capacitance,
$C = \frac{ε_{0} A}{d}$
Area $(a) = A / 2$
Diameter $(d) = 2 d$
Then, $C^{'} = \frac{ε_{0} A}{2 \times 2 d}$
$C^{'} = \frac{ε_{0} A}{4 d} = \frac{C}{4}$
$C^{'} = \frac{C}{4}$

Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$ , where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $C_{0} =$ capacitance of capacitor with air as medium between plates.)

1

\frac{4 {KC}_{0}}{3 + K}

2

\frac{3 {KC}_{0}}{3 + K}

3

\frac{3 + K}{4 {KC}_{0}}

4

\frac{K}{4 + K}

Explanation:

: Given that,
Thickness of slab $(t) = \frac{3}{4} d$
The capacitance of a parallel plate capacitor in the absence of the dielectric is-
$C_{0} = \frac{ε_{0} A}{d}$
If a dielectric slab is partially filled between the plates of capacitor, then
$C = \frac{ε_{0} A}{[d - t + \frac{t}{K}]}$
$C = \frac{ε_{0} A}{(d - \frac{3}{4} d) + (\frac{3 d}{4 K})}$
$C = \frac{ε_{0} A}{\frac{d}{4} + \frac{3 d}{4 K}}$
$C = \frac{4 K ε_{0} A}{d (K + 3)}$
$C = \frac{4 K ε_{0} A}{d (K + 3)} \times \frac{d}{ε_{0} A}$
$C = \frac{4 KC C_{0}}{(3 + K)}$

[JEE Main-28.07.2022

Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $C_{1} / C_{2}$ is

1

\frac{1}{4}

2

\frac{1}{16}

3

\frac{1}{8}

4

\frac{1}{12}

Explanation:

: Let capacitance of each capacitor is c In series, net capacitance,
$C_{2} = \frac{C}{4}$
In Parallel, net capacitance
$C_{2} = 4 C$
The ratio of $\frac{C_{1}}{C_{2}}$ is,
$\frac{C_{1}}{C_{2}} = \frac{C}{4 \times 4 C} = \frac{1}{16}$
$\frac{C_{1}}{C_{2}} = \frac{1}{16}$

[WB JEE-2007]

Capacitance

165663 The effective capacitances of two capacitors are $3 μ F$ and $16 μ F$ , when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1

10 μ F, 6 μ F

2

8 μ F, 8 μ F

3

12 μ F, 4 μ F

4

1.2 μ F, 1.8 μ F

Explanation:

: Given,
$C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = 3 μ F$
$(C_{s}$ is equivalent capacitance of capacitor in series)
$C_{p} = C_{1} + C_{2} = 16 μ F$
$(C_{p}$ is equivalent capacitance of capacitor in Parallel)
From (i) \& (ii)
$C_{1} C_{2} = 3 \times 16 = 48$
From (ii) \& (iii)
From (iii)
$C_{1} + \frac{48}{C_{1}} = 16$
$C_{1}^{2} - 16 C_{1} + 48 = 0$
$C_{1}^{2} - 12 C_{1} - 4 C_{1} + 48 = 0$
$C_{1} (C_{1} - 12) - 4 (C_{1} - 12) = 0$
$(C_{1} - 12) (C_{1} - 4) = 0$
$C_{1} = 12 μ F, 4 μ F$
$12 \times C_{2} = 48$
$C_{2} = \frac{48}{12}$
$= 4 μ F$
$C_{1} = 12 μ F C_{2} = 4 μ F$

Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1

2 C

2

C / 2

3

4 C

4

C / 4

Explanation:

: Capacitance,
$C = \frac{ε_{0} A}{d}$
Area $(a) = A / 2$
Diameter $(d) = 2 d$
Then, $C^{'} = \frac{ε_{0} A}{2 \times 2 d}$
$C^{'} = \frac{ε_{0} A}{4 d} = \frac{C}{4}$
$C^{'} = \frac{C}{4}$

Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$ , where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $C_{0} =$ capacitance of capacitor with air as medium between plates.)

1

\frac{4 {KC}_{0}}{3 + K}

2

\frac{3 {KC}_{0}}{3 + K}

3

\frac{3 + K}{4 {KC}_{0}}

4

\frac{K}{4 + K}

Explanation:

: Given that,
Thickness of slab $(t) = \frac{3}{4} d$
The capacitance of a parallel plate capacitor in the absence of the dielectric is-
$C_{0} = \frac{ε_{0} A}{d}$
If a dielectric slab is partially filled between the plates of capacitor, then
$C = \frac{ε_{0} A}{[d - t + \frac{t}{K}]}$
$C = \frac{ε_{0} A}{(d - \frac{3}{4} d) + (\frac{3 d}{4 K})}$
$C = \frac{ε_{0} A}{\frac{d}{4} + \frac{3 d}{4 K}}$
$C = \frac{4 K ε_{0} A}{d (K + 3)}$
$C = \frac{4 K ε_{0} A}{d (K + 3)} \times \frac{d}{ε_{0} A}$
$C = \frac{4 KC C_{0}}{(3 + K)}$

[JEE Main-28.07.2022

Capacitance

165662 Four capacitors of equal capacitance have a net capacitance $C_{1}$ when connected in series and a net capacitance $C_{2}$ when connected in parallel. The ratio of $C_{1} / C_{2}$ is

1

\frac{1}{4}

2

\frac{1}{16}

3

\frac{1}{8}

4

\frac{1}{12}

Explanation:

: Let capacitance of each capacitor is c In series, net capacitance,
$C_{2} = \frac{C}{4}$
In Parallel, net capacitance
$C_{2} = 4 C$
The ratio of $\frac{C_{1}}{C_{2}}$ is,
$\frac{C_{1}}{C_{2}} = \frac{C}{4 \times 4 C} = \frac{1}{16}$
$\frac{C_{1}}{C_{2}} = \frac{1}{16}$

[WB JEE-2007]

Capacitance

165663 The effective capacitances of two capacitors are $3 μ F$ and $16 μ F$ , when they are connected in series and parallel respectively. The capacitance of two capacitors are:

1

10 μ F, 6 μ F

2

8 μ F, 8 μ F

3

12 μ F, 4 μ F

4

1.2 μ F, 1.8 μ F

Explanation:

: Given,
$C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = 3 μ F$
$(C_{s}$ is equivalent capacitance of capacitor in series)
$C_{p} = C_{1} + C_{2} = 16 μ F$
$(C_{p}$ is equivalent capacitance of capacitor in Parallel)
From (i) \& (ii)
$C_{1} C_{2} = 3 \times 16 = 48$
From (ii) \& (iii)
From (iii)
$C_{1} + \frac{48}{C_{1}} = 16$
$C_{1}^{2} - 16 C_{1} + 48 = 0$
$C_{1}^{2} - 12 C_{1} - 4 C_{1} + 48 = 0$
$C_{1} (C_{1} - 12) - 4 (C_{1} - 12) = 0$
$(C_{1} - 12) (C_{1} - 4) = 0$
$C_{1} = 12 μ F, 4 μ F$
$12 \times C_{2} = 48$
$C_{2} = \frac{48}{12}$
$= 4 μ F$
$C_{1} = 12 μ F C_{2} = 4 μ F$

Capacitance

165664 The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If $C$ is initial capacitance, its final capacitance is equal to:

1

2 C

2

C / 2

3

4 C

4

C / 4

Explanation:

: Capacitance,
$C = \frac{ε_{0} A}{d}$
Area $(a) = A / 2$
Diameter $(d) = 2 d$
Then, $C^{'} = \frac{ε_{0} A}{2 \times 2 d}$
$C^{'} = \frac{ε_{0} A}{4 d} = \frac{C}{4}$
$C^{'} = \frac{C}{4}$

Capacitance

165665 A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} d$ , where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $C_{0} =$ capacitance of capacitor with air as medium between plates.)

1

\frac{4 {KC}_{0}}{3 + K}

2

\frac{3 {KC}_{0}}{3 + K}

3

\frac{3 + K}{4 {KC}_{0}}

4

\frac{K}{4 + K}

Explanation:

: Given that,
Thickness of slab $(t) = \frac{3}{4} d$
The capacitance of a parallel plate capacitor in the absence of the dielectric is-
$C_{0} = \frac{ε_{0} A}{d}$
If a dielectric slab is partially filled between the plates of capacitor, then
$C = \frac{ε_{0} A}{[d - t + \frac{t}{K}]}$
$C = \frac{ε_{0} A}{(d - \frac{3}{4} d) + (\frac{3 d}{4 K})}$
$C = \frac{ε_{0} A}{\frac{d}{4} + \frac{3 d}{4 K}}$
$C = \frac{4 K ε_{0} A}{d (K + 3)}$
$C = \frac{4 K ε_{0} A}{d (K + 3)} \times \frac{d}{ε_{0} A}$
$C = \frac{4 KC C_{0}}{(3 + K)}$

[JEE Main-28.07.2022

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