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Capacitance

165680 Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

1 1.5

2 2.5

3 1.2

4 3

Explanation:

Given, $C = 40 μ F$
Capacitor is filled with dielectric slob of $K$ .
Thus, equivalent capacitance,
$C_{eq} = \frac{C (KC)}{C + (KC)} = \frac{{KC}^{2}}{C (1 + K)} = \frac{KC}{(1 + K)}$
$\frac{KC}{1 + K} = 24$
$KC = 24 + 24 K$
$40 K = 24 + 24 K$
$40 K - 24 K = 24$
$16 K = 24$
$K = \frac{24}{16}$
$K = 1.5$

[JEE Main-28.07.2022

Capacitance

165681 The capacitance of spherical conductor with radius $1 m$ is

1

9 \times 10^{- 9} F

2

10 μ F

3

2.5 \times 10^{- 10} F

4

1.1 \times 10^{- 10} F

Explanation:

: Given,
$r =$ Radius of sphere $= 1 m$
Capacitance of spherical conductor
$C = 4 π ε_{0} r$
We know that,
$\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$
$4 π ε_{0} = \frac{1}{9 \times 10^{9}}$
$∴ C = \frac{1}{9 \times 10^{9}} = \frac{1 \times 10^{- 9}}{9} = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[UP CPMT-2012]

Capacitance

165682 An electric circuit requires a total capacitance of $2 μ F$ across a potential of $1000 V$ . Large number of $1 μ F$ capacitances are available each of which would breakdown if the potential is more than $350 V$ . How many capacitances are required to make the circuit?

1 24

2 20

3 18

4 12

Explanation:

: Here,
Capacitance of each capacitor
$C = 1 μ F$
Voltage rating of each capacitor $= 350 V$
Supply voltage $= 1000 V$
Total capacitance $= 2 μ F$
Let, $n$ capacitors of $1 μ F$ each be connected in series in a row and $m$ such rows be connected in parallel as shown in the figure.

$∵$ Each capacitor can withstand $350 V$ .
$∴ n = \frac{1000}{350} = 2.8$
As $n$ cannot be fraction, therefore $n = 3$
Capacitance of each row of 3 capacitors of $1 μ F$ each in series is-
$C_{s} = \frac{1}{3} μ F$
$∴ \frac{m}{3} μ F = 2 μ F$
$Or m = 6$
Total number of capacitors $= n \times m$
$= 3 \times 6$
$= 18$

[AMU-2012]

Capacitance

165683 Find the total capacitance and total charge on the capacitors

1

1.5 nF, 9 nC

2

3.0 nF, 18 nC

3

1.5 nF, 4.5 nC

4

3.0 nF, 9 nC

5

3.0 nF, 4.5 nC

Explanation:

: Given diagram,

$1 nF & 2 nF$ are in parallel \& equivalent of these two are in series with $3 nF$ ,
$∴ C_{eq} = (1 + 2) nF$ series with $3 nF$
$C_{eq} = 3 nF$ series with $3 nF$
$∴ C_{eq} = \frac{3 \times 3}{3 + 3} = 1.5 nF$
$C_{eq} = 1.5 nF$
Change on circuit
$Q = C_{eq} V = 1.5 \times 6$
$= 9.0 nC$
$Q = 9 nC$

[Kerala CEE-2019]

Capacitance

165680 Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

1 1.5

2 2.5

3 1.2

4 3

Explanation:

Given, $C = 40 μ F$
Capacitor is filled with dielectric slob of $K$ .
Thus, equivalent capacitance,
$C_{eq} = \frac{C (KC)}{C + (KC)} = \frac{{KC}^{2}}{C (1 + K)} = \frac{KC}{(1 + K)}$
$\frac{KC}{1 + K} = 24$
$KC = 24 + 24 K$
$40 K = 24 + 24 K$
$40 K - 24 K = 24$
$16 K = 24$
$K = \frac{24}{16}$
$K = 1.5$

[JEE Main-28.07.2022

Capacitance

165681 The capacitance of spherical conductor with radius $1 m$ is

1

9 \times 10^{- 9} F

2

10 μ F

3

2.5 \times 10^{- 10} F

4

1.1 \times 10^{- 10} F

Explanation:

: Given,
$r =$ Radius of sphere $= 1 m$
Capacitance of spherical conductor
$C = 4 π ε_{0} r$
We know that,
$\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$
$4 π ε_{0} = \frac{1}{9 \times 10^{9}}$
$∴ C = \frac{1}{9 \times 10^{9}} = \frac{1 \times 10^{- 9}}{9} = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[UP CPMT-2012]

Capacitance

165682 An electric circuit requires a total capacitance of $2 μ F$ across a potential of $1000 V$ . Large number of $1 μ F$ capacitances are available each of which would breakdown if the potential is more than $350 V$ . How many capacitances are required to make the circuit?

1 24

2 20

3 18

4 12

Explanation:

: Here,
Capacitance of each capacitor
$C = 1 μ F$
Voltage rating of each capacitor $= 350 V$
Supply voltage $= 1000 V$
Total capacitance $= 2 μ F$
Let, $n$ capacitors of $1 μ F$ each be connected in series in a row and $m$ such rows be connected in parallel as shown in the figure.

$∵$ Each capacitor can withstand $350 V$ .
$∴ n = \frac{1000}{350} = 2.8$
As $n$ cannot be fraction, therefore $n = 3$
Capacitance of each row of 3 capacitors of $1 μ F$ each in series is-
$C_{s} = \frac{1}{3} μ F$
$∴ \frac{m}{3} μ F = 2 μ F$
$Or m = 6$
Total number of capacitors $= n \times m$
$= 3 \times 6$
$= 18$

[AMU-2012]

Capacitance

165683 Find the total capacitance and total charge on the capacitors

1

1.5 nF, 9 nC

2

3.0 nF, 18 nC

3

1.5 nF, 4.5 nC

4

3.0 nF, 9 nC

5

3.0 nF, 4.5 nC

Explanation:

: Given diagram,

$1 nF & 2 nF$ are in parallel \& equivalent of these two are in series with $3 nF$ ,
$∴ C_{eq} = (1 + 2) nF$ series with $3 nF$
$C_{eq} = 3 nF$ series with $3 nF$
$∴ C_{eq} = \frac{3 \times 3}{3 + 3} = 1.5 nF$
$C_{eq} = 1.5 nF$
Change on circuit
$Q = C_{eq} V = 1.5 \times 6$
$= 9.0 nC$
$Q = 9 nC$

[Kerala CEE-2019]

Capacitance

165680 Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

1 1.5

2 2.5

3 1.2

4 3

Explanation:

Given, $C = 40 μ F$
Capacitor is filled with dielectric slob of $K$ .
Thus, equivalent capacitance,
$C_{eq} = \frac{C (KC)}{C + (KC)} = \frac{{KC}^{2}}{C (1 + K)} = \frac{KC}{(1 + K)}$
$\frac{KC}{1 + K} = 24$
$KC = 24 + 24 K$
$40 K = 24 + 24 K$
$40 K - 24 K = 24$
$16 K = 24$
$K = \frac{24}{16}$
$K = 1.5$

[JEE Main-28.07.2022

Capacitance

165681 The capacitance of spherical conductor with radius $1 m$ is

1

9 \times 10^{- 9} F

2

10 μ F

3

2.5 \times 10^{- 10} F

4

1.1 \times 10^{- 10} F

Explanation:

: Given,
$r =$ Radius of sphere $= 1 m$
Capacitance of spherical conductor
$C = 4 π ε_{0} r$
We know that,
$\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$
$4 π ε_{0} = \frac{1}{9 \times 10^{9}}$
$∴ C = \frac{1}{9 \times 10^{9}} = \frac{1 \times 10^{- 9}}{9} = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[UP CPMT-2012]

Capacitance

165682 An electric circuit requires a total capacitance of $2 μ F$ across a potential of $1000 V$ . Large number of $1 μ F$ capacitances are available each of which would breakdown if the potential is more than $350 V$ . How many capacitances are required to make the circuit?

1 24

2 20

3 18

4 12

Explanation:

: Here,
Capacitance of each capacitor
$C = 1 μ F$
Voltage rating of each capacitor $= 350 V$
Supply voltage $= 1000 V$
Total capacitance $= 2 μ F$
Let, $n$ capacitors of $1 μ F$ each be connected in series in a row and $m$ such rows be connected in parallel as shown in the figure.

$∵$ Each capacitor can withstand $350 V$ .
$∴ n = \frac{1000}{350} = 2.8$
As $n$ cannot be fraction, therefore $n = 3$
Capacitance of each row of 3 capacitors of $1 μ F$ each in series is-
$C_{s} = \frac{1}{3} μ F$
$∴ \frac{m}{3} μ F = 2 μ F$
$Or m = 6$
Total number of capacitors $= n \times m$
$= 3 \times 6$
$= 18$

[AMU-2012]

Capacitance

165683 Find the total capacitance and total charge on the capacitors

1

1.5 nF, 9 nC

2

3.0 nF, 18 nC

3

1.5 nF, 4.5 nC

4

3.0 nF, 9 nC

5

3.0 nF, 4.5 nC

Explanation:

: Given diagram,

$1 nF & 2 nF$ are in parallel \& equivalent of these two are in series with $3 nF$ ,
$∴ C_{eq} = (1 + 2) nF$ series with $3 nF$
$C_{eq} = 3 nF$ series with $3 nF$
$∴ C_{eq} = \frac{3 \times 3}{3 + 3} = 1.5 nF$
$C_{eq} = 1.5 nF$
Change on circuit
$Q = C_{eq} V = 1.5 \times 6$
$= 9.0 nC$
$Q = 9 nC$

[Kerala CEE-2019]

Capacitance

165680 Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24 μ F$ . The value of $K$ will be :

1 1.5

2 2.5

3 1.2

4 3

Explanation:

Given, $C = 40 μ F$
Capacitor is filled with dielectric slob of $K$ .
Thus, equivalent capacitance,
$C_{eq} = \frac{C (KC)}{C + (KC)} = \frac{{KC}^{2}}{C (1 + K)} = \frac{KC}{(1 + K)}$
$\frac{KC}{1 + K} = 24$
$KC = 24 + 24 K$
$40 K = 24 + 24 K$
$40 K - 24 K = 24$
$16 K = 24$
$K = \frac{24}{16}$
$K = 1.5$

[JEE Main-28.07.2022

Capacitance

165681 The capacitance of spherical conductor with radius $1 m$ is

1

9 \times 10^{- 9} F

2

10 μ F

3

2.5 \times 10^{- 10} F

4

1.1 \times 10^{- 10} F

Explanation:

: Given,
$r =$ Radius of sphere $= 1 m$
Capacitance of spherical conductor
$C = 4 π ε_{0} r$
We know that,
$\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$
$4 π ε_{0} = \frac{1}{9 \times 10^{9}}$
$∴ C = \frac{1}{9 \times 10^{9}} = \frac{1 \times 10^{- 9}}{9} = 0.11 \times 10^{- 9} F$
$C = 1.1 \times 10^{- 10} F$

[UP CPMT-2012]

Capacitance

165682 An electric circuit requires a total capacitance of $2 μ F$ across a potential of $1000 V$ . Large number of $1 μ F$ capacitances are available each of which would breakdown if the potential is more than $350 V$ . How many capacitances are required to make the circuit?

1 24

2 20

3 18

4 12

Explanation:

: Here,
Capacitance of each capacitor
$C = 1 μ F$
Voltage rating of each capacitor $= 350 V$
Supply voltage $= 1000 V$
Total capacitance $= 2 μ F$
Let, $n$ capacitors of $1 μ F$ each be connected in series in a row and $m$ such rows be connected in parallel as shown in the figure.

$∵$ Each capacitor can withstand $350 V$ .
$∴ n = \frac{1000}{350} = 2.8$
As $n$ cannot be fraction, therefore $n = 3$
Capacitance of each row of 3 capacitors of $1 μ F$ each in series is-
$C_{s} = \frac{1}{3} μ F$
$∴ \frac{m}{3} μ F = 2 μ F$
$Or m = 6$
Total number of capacitors $= n \times m$
$= 3 \times 6$
$= 18$

[AMU-2012]

Capacitance

165683 Find the total capacitance and total charge on the capacitors

1

1.5 nF, 9 nC

2

3.0 nF, 18 nC

3

1.5 nF, 4.5 nC

4

3.0 nF, 9 nC

5

3.0 nF, 4.5 nC

Explanation:

: Given diagram,

$1 nF & 2 nF$ are in parallel \& equivalent of these two are in series with $3 nF$ ,
$∴ C_{eq} = (1 + 2) nF$ series with $3 nF$
$C_{eq} = 3 nF$ series with $3 nF$
$∴ C_{eq} = \frac{3 \times 3}{3 + 3} = 1.5 nF$
$C_{eq} = 1.5 nF$
Change on circuit
$Q = C_{eq} V = 1.5 \times 6$
$= 9.0 nC$
$Q = 9 nC$

[Kerala CEE-2019]