QBManager

Capacitance

165622 A potential difference of $300 V$ is applied to a combination of $2.0 μ F$ and $8.0 μ F$ capacitors connected in series. The charge on the $2.0 μ F$ capacitor is

1

2.4 \times 10^{- 4} C

2

4.8 \times 10^{- 4} C

3

7.2 \times 10^{- 4} C

4

9.6 \times 10^{- 4} C

Explanation:

: Potential difference $(V) = 300 V$

In series, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$
$C = \frac{8}{5} μ F = \frac{8}{5} \times 10^{- 6} F$
Charge $(Q) = VC$
$= 300 \times \frac{8}{5} \times 10^{- 6}$
$= 4.8 \times 10^{- 4} C$

[VITEEE-2006]

Capacitance

165623 Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y (A \to$ Area of each plate, $d \to$ distance between the plates) is :

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:

:

Capacitance of each capacitor,

Capacitor across $V$ and $Y$ are in parallel,
$C_{1} = C + C = 2 C$
$∵$ Capacitor $C$ and $C_{1}$ will be in series,
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C} = \frac{1}{2 C} + \frac{1}{C} = \frac{3}{2 C}$
$C_{eq} = \frac{2 C}{3}$
$C_{eq} = \frac{2}{3} \frac{ε_{0} A}{d}$

[UP CPMT-2012

Capacitance

165624 A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when:

1 an iron rod is introduced into the inductance coil

2 the number of turns in the inductance coil is increased

3 separation between the plates of the capacitor is increased

4 a dielectric is introduced into the gap between the plates of the capacitor

Explanation:

:

Resistance of bulb $= R$
$∵$ Current (i) $= \frac{V}{\sqrt{R^{2} + C^{2}}}$
If a dielectric is introduced into the gap between the plates, of capacitor, its capacitance will increase and hence impedance of the circuit will decrease.
Thus, current and hence brightness of the bulb increase

[Karnataka CET-2010]

Capacitance

165625 In a parallel plate capacitor of capacitance $C$ , a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes :

1

C / 4

2

C / 2

3

2 C

4

4 C

Explanation:

: We know that,
$∵ C = \frac{ε_{o} A}{d}$
With dielectric $k$ is filled partially between plates,
$C_{1} = \frac{ε_{o} A}{d - t + t / k}$
If a dielectric slab of dielectric constant $k$ and thickness $t$ is inserted in the air gap of a capacitor of plate separation $d$ and plate area $A$ , its capacitance becomes. For metal sheet,
$t = \frac{d}{2}, k = \infty$
$∴ C_{2} = \frac{ε_{0} A}{d - t}$
$= \frac{ε_{0} A}{d - \frac{d}{2}} [∵ t = \frac{d}{2}]$
$C_{2} = 2 \frac{ε_{0} A}{d}$
$C_{2} = 2 C$

[Karnataka CET-2001]

Capacitance

165627 $n$ identical capacitors each of capacitance $C$ when connected in parallel give the effective capacitance $90 μ F$ and when connected in series give $2.5 μ F$ . Then the values of $n$ and $C$ respectively are

1 6 and

15 μ F

2 5 and

18 μ F

3 15 and

6 μ F

4 18 and

5 μ F

Explanation:

:Given that,
When connected in parallel
${(C_{eq})}_{1} = 90 μ F$
' $n$ ' identical capacitance in parallel,
$(C_{eq}) = C_{1} + C_{2} + \dots \dots . . C_{n}$
$(C_{eq}) = nC \Rightarrow 90 = nC$
$n = \frac{90}{C}$
When ' $n$ ' identical capacitor connected in series,
$\frac{1}{C_{e q}} = \frac{n}{C}$
$C_{eq} = \frac{C}{n}$
$2.5 μ F = \frac{C}{n}$
$C = 2.5 \times n$
$C = 2.5 \times \frac{90}{C} [from equation(ii)]$
$C^{2} = 225$
$C = 15 μ F$
$C = 15 μ F in equation (i) we get -$
$n \times 15 = 90$
$n = 6$
Putting $C = 15 μ F$ in equation (i) we get -

[J and K CET- 2010]

Capacitance

165622 A potential difference of $300 V$ is applied to a combination of $2.0 μ F$ and $8.0 μ F$ capacitors connected in series. The charge on the $2.0 μ F$ capacitor is

1

2.4 \times 10^{- 4} C

2

4.8 \times 10^{- 4} C

3

7.2 \times 10^{- 4} C

4

9.6 \times 10^{- 4} C

Explanation:

: Potential difference $(V) = 300 V$

In series, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$
$C = \frac{8}{5} μ F = \frac{8}{5} \times 10^{- 6} F$
Charge $(Q) = VC$
$= 300 \times \frac{8}{5} \times 10^{- 6}$
$= 4.8 \times 10^{- 4} C$

[VITEEE-2006]

Capacitance

165623 Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y (A \to$ Area of each plate, $d \to$ distance between the plates) is :

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:

:

Capacitance of each capacitor,

Capacitor across $V$ and $Y$ are in parallel,
$C_{1} = C + C = 2 C$
$∵$ Capacitor $C$ and $C_{1}$ will be in series,
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C} = \frac{1}{2 C} + \frac{1}{C} = \frac{3}{2 C}$
$C_{eq} = \frac{2 C}{3}$
$C_{eq} = \frac{2}{3} \frac{ε_{0} A}{d}$

[UP CPMT-2012

Capacitance

165624 A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when:

1 an iron rod is introduced into the inductance coil

2 the number of turns in the inductance coil is increased

3 separation between the plates of the capacitor is increased

4 a dielectric is introduced into the gap between the plates of the capacitor

Explanation:

:

Resistance of bulb $= R$
$∵$ Current (i) $= \frac{V}{\sqrt{R^{2} + C^{2}}}$
If a dielectric is introduced into the gap between the plates, of capacitor, its capacitance will increase and hence impedance of the circuit will decrease.
Thus, current and hence brightness of the bulb increase

[Karnataka CET-2010]

Capacitance

165625 In a parallel plate capacitor of capacitance $C$ , a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes :

1

C / 4

2

C / 2

3

2 C

4

4 C

Explanation:

: We know that,
$∵ C = \frac{ε_{o} A}{d}$
With dielectric $k$ is filled partially between plates,
$C_{1} = \frac{ε_{o} A}{d - t + t / k}$
If a dielectric slab of dielectric constant $k$ and thickness $t$ is inserted in the air gap of a capacitor of plate separation $d$ and plate area $A$ , its capacitance becomes. For metal sheet,
$t = \frac{d}{2}, k = \infty$
$∴ C_{2} = \frac{ε_{0} A}{d - t}$
$= \frac{ε_{0} A}{d - \frac{d}{2}} [∵ t = \frac{d}{2}]$
$C_{2} = 2 \frac{ε_{0} A}{d}$
$C_{2} = 2 C$

[Karnataka CET-2001]

Capacitance

165627 $n$ identical capacitors each of capacitance $C$ when connected in parallel give the effective capacitance $90 μ F$ and when connected in series give $2.5 μ F$ . Then the values of $n$ and $C$ respectively are

1 6 and

15 μ F

2 5 and

18 μ F

3 15 and

6 μ F

4 18 and

5 μ F

Explanation:

:Given that,
When connected in parallel
${(C_{eq})}_{1} = 90 μ F$
' $n$ ' identical capacitance in parallel,
$(C_{eq}) = C_{1} + C_{2} + \dots \dots . . C_{n}$
$(C_{eq}) = nC \Rightarrow 90 = nC$
$n = \frac{90}{C}$
When ' $n$ ' identical capacitor connected in series,
$\frac{1}{C_{e q}} = \frac{n}{C}$
$C_{eq} = \frac{C}{n}$
$2.5 μ F = \frac{C}{n}$
$C = 2.5 \times n$
$C = 2.5 \times \frac{90}{C} [from equation(ii)]$
$C^{2} = 225$
$C = 15 μ F$
$C = 15 μ F in equation (i) we get -$
$n \times 15 = 90$
$n = 6$
Putting $C = 15 μ F$ in equation (i) we get -

[J and K CET- 2010]

Capacitance

165622 A potential difference of $300 V$ is applied to a combination of $2.0 μ F$ and $8.0 μ F$ capacitors connected in series. The charge on the $2.0 μ F$ capacitor is

1

2.4 \times 10^{- 4} C

2

4.8 \times 10^{- 4} C

3

7.2 \times 10^{- 4} C

4

9.6 \times 10^{- 4} C

Explanation:

: Potential difference $(V) = 300 V$

In series, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$
$C = \frac{8}{5} μ F = \frac{8}{5} \times 10^{- 6} F$
Charge $(Q) = VC$
$= 300 \times \frac{8}{5} \times 10^{- 6}$
$= 4.8 \times 10^{- 4} C$

[VITEEE-2006]

Capacitance

165623 Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y (A \to$ Area of each plate, $d \to$ distance between the plates) is :

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:

:

Capacitance of each capacitor,

Capacitor across $V$ and $Y$ are in parallel,
$C_{1} = C + C = 2 C$
$∵$ Capacitor $C$ and $C_{1}$ will be in series,
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C} = \frac{1}{2 C} + \frac{1}{C} = \frac{3}{2 C}$
$C_{eq} = \frac{2 C}{3}$
$C_{eq} = \frac{2}{3} \frac{ε_{0} A}{d}$

[UP CPMT-2012

Capacitance

165624 A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when:

1 an iron rod is introduced into the inductance coil

2 the number of turns in the inductance coil is increased

3 separation between the plates of the capacitor is increased

4 a dielectric is introduced into the gap between the plates of the capacitor

Explanation:

:

Resistance of bulb $= R$
$∵$ Current (i) $= \frac{V}{\sqrt{R^{2} + C^{2}}}$
If a dielectric is introduced into the gap between the plates, of capacitor, its capacitance will increase and hence impedance of the circuit will decrease.
Thus, current and hence brightness of the bulb increase

[Karnataka CET-2010]

Capacitance

165625 In a parallel plate capacitor of capacitance $C$ , a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes :

1

C / 4

2

C / 2

3

2 C

4

4 C

Explanation:

: We know that,
$∵ C = \frac{ε_{o} A}{d}$
With dielectric $k$ is filled partially between plates,
$C_{1} = \frac{ε_{o} A}{d - t + t / k}$
If a dielectric slab of dielectric constant $k$ and thickness $t$ is inserted in the air gap of a capacitor of plate separation $d$ and plate area $A$ , its capacitance becomes. For metal sheet,
$t = \frac{d}{2}, k = \infty$
$∴ C_{2} = \frac{ε_{0} A}{d - t}$
$= \frac{ε_{0} A}{d - \frac{d}{2}} [∵ t = \frac{d}{2}]$
$C_{2} = 2 \frac{ε_{0} A}{d}$
$C_{2} = 2 C$

[Karnataka CET-2001]

Capacitance

165627 $n$ identical capacitors each of capacitance $C$ when connected in parallel give the effective capacitance $90 μ F$ and when connected in series give $2.5 μ F$ . Then the values of $n$ and $C$ respectively are

1 6 and

15 μ F

2 5 and

18 μ F

3 15 and

6 μ F

4 18 and

5 μ F

Explanation:

:Given that,
When connected in parallel
${(C_{eq})}_{1} = 90 μ F$
' $n$ ' identical capacitance in parallel,
$(C_{eq}) = C_{1} + C_{2} + \dots \dots . . C_{n}$
$(C_{eq}) = nC \Rightarrow 90 = nC$
$n = \frac{90}{C}$
When ' $n$ ' identical capacitor connected in series,
$\frac{1}{C_{e q}} = \frac{n}{C}$
$C_{eq} = \frac{C}{n}$
$2.5 μ F = \frac{C}{n}$
$C = 2.5 \times n$
$C = 2.5 \times \frac{90}{C} [from equation(ii)]$
$C^{2} = 225$
$C = 15 μ F$
$C = 15 μ F in equation (i) we get -$
$n \times 15 = 90$
$n = 6$
Putting $C = 15 μ F$ in equation (i) we get -

[J and K CET- 2010]

Capacitance

165622 A potential difference of $300 V$ is applied to a combination of $2.0 μ F$ and $8.0 μ F$ capacitors connected in series. The charge on the $2.0 μ F$ capacitor is

1

2.4 \times 10^{- 4} C

2

4.8 \times 10^{- 4} C

3

7.2 \times 10^{- 4} C

4

9.6 \times 10^{- 4} C

Explanation:

: Potential difference $(V) = 300 V$

In series, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$
$C = \frac{8}{5} μ F = \frac{8}{5} \times 10^{- 6} F$
Charge $(Q) = VC$
$= 300 \times \frac{8}{5} \times 10^{- 6}$
$= 4.8 \times 10^{- 4} C$

[VITEEE-2006]

Capacitance

165623 Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y (A \to$ Area of each plate, $d \to$ distance between the plates) is :

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:

:

Capacitance of each capacitor,

Capacitor across $V$ and $Y$ are in parallel,
$C_{1} = C + C = 2 C$
$∵$ Capacitor $C$ and $C_{1}$ will be in series,
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C} = \frac{1}{2 C} + \frac{1}{C} = \frac{3}{2 C}$
$C_{eq} = \frac{2 C}{3}$
$C_{eq} = \frac{2}{3} \frac{ε_{0} A}{d}$

[UP CPMT-2012

Capacitance

165624 A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when:

1 an iron rod is introduced into the inductance coil

2 the number of turns in the inductance coil is increased

3 separation between the plates of the capacitor is increased

4 a dielectric is introduced into the gap between the plates of the capacitor

Explanation:

:

Resistance of bulb $= R$
$∵$ Current (i) $= \frac{V}{\sqrt{R^{2} + C^{2}}}$
If a dielectric is introduced into the gap between the plates, of capacitor, its capacitance will increase and hence impedance of the circuit will decrease.
Thus, current and hence brightness of the bulb increase

[Karnataka CET-2010]

Capacitance

165625 In a parallel plate capacitor of capacitance $C$ , a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes :

1

C / 4

2

C / 2

3

2 C

4

4 C

Explanation:

: We know that,
$∵ C = \frac{ε_{o} A}{d}$
With dielectric $k$ is filled partially between plates,
$C_{1} = \frac{ε_{o} A}{d - t + t / k}$
If a dielectric slab of dielectric constant $k$ and thickness $t$ is inserted in the air gap of a capacitor of plate separation $d$ and plate area $A$ , its capacitance becomes. For metal sheet,
$t = \frac{d}{2}, k = \infty$
$∴ C_{2} = \frac{ε_{0} A}{d - t}$
$= \frac{ε_{0} A}{d - \frac{d}{2}} [∵ t = \frac{d}{2}]$
$C_{2} = 2 \frac{ε_{0} A}{d}$
$C_{2} = 2 C$

[Karnataka CET-2001]

Capacitance

165627 $n$ identical capacitors each of capacitance $C$ when connected in parallel give the effective capacitance $90 μ F$ and when connected in series give $2.5 μ F$ . Then the values of $n$ and $C$ respectively are

1 6 and

15 μ F

2 5 and

18 μ F

3 15 and

6 μ F

4 18 and

5 μ F

Explanation:

:Given that,
When connected in parallel
${(C_{eq})}_{1} = 90 μ F$
' $n$ ' identical capacitance in parallel,
$(C_{eq}) = C_{1} + C_{2} + \dots \dots . . C_{n}$
$(C_{eq}) = nC \Rightarrow 90 = nC$
$n = \frac{90}{C}$
When ' $n$ ' identical capacitor connected in series,
$\frac{1}{C_{e q}} = \frac{n}{C}$
$C_{eq} = \frac{C}{n}$
$2.5 μ F = \frac{C}{n}$
$C = 2.5 \times n$
$C = 2.5 \times \frac{90}{C} [from equation(ii)]$
$C^{2} = 225$
$C = 15 μ F$
$C = 15 μ F in equation (i) we get -$
$n \times 15 = 90$
$n = 6$
Putting $C = 15 μ F$ in equation (i) we get -

[J and K CET- 2010]

Capacitance

165622 A potential difference of $300 V$ is applied to a combination of $2.0 μ F$ and $8.0 μ F$ capacitors connected in series. The charge on the $2.0 μ F$ capacitor is

1

2.4 \times 10^{- 4} C

2

4.8 \times 10^{- 4} C

3

7.2 \times 10^{- 4} C

4

9.6 \times 10^{- 4} C

Explanation:

: Potential difference $(V) = 300 V$

In series, $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$
$C = \frac{8}{5} μ F = \frac{8}{5} \times 10^{- 6} F$
Charge $(Q) = VC$
$= 300 \times \frac{8}{5} \times 10^{- 6}$
$= 4.8 \times 10^{- 4} C$

[VITEEE-2006]

Capacitance

165623 Four metal plates are arranged as shown in the figure. Capacitance between $X$ and $Y (A \to$ Area of each plate, $d \to$ distance between the plates) is :

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:

:

Capacitance of each capacitor,

Capacitor across $V$ and $Y$ are in parallel,
$C_{1} = C + C = 2 C$
$∵$ Capacitor $C$ and $C_{1}$ will be in series,
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C} = \frac{1}{2 C} + \frac{1}{C} = \frac{3}{2 C}$
$C_{eq} = \frac{2 C}{3}$
$C_{eq} = \frac{2}{3} \frac{ε_{0} A}{d}$

[UP CPMT-2012

Capacitance

165624 A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when:

1 an iron rod is introduced into the inductance coil

2 the number of turns in the inductance coil is increased

3 separation between the plates of the capacitor is increased

4 a dielectric is introduced into the gap between the plates of the capacitor

Explanation:

:

Resistance of bulb $= R$
$∵$ Current (i) $= \frac{V}{\sqrt{R^{2} + C^{2}}}$
If a dielectric is introduced into the gap between the plates, of capacitor, its capacitance will increase and hence impedance of the circuit will decrease.
Thus, current and hence brightness of the bulb increase

[Karnataka CET-2010]

Capacitance

165625 In a parallel plate capacitor of capacitance $C$ , a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes :

1

C / 4

2

C / 2

3

2 C

4

4 C

Explanation:

: We know that,
$∵ C = \frac{ε_{o} A}{d}$
With dielectric $k$ is filled partially between plates,
$C_{1} = \frac{ε_{o} A}{d - t + t / k}$
If a dielectric slab of dielectric constant $k$ and thickness $t$ is inserted in the air gap of a capacitor of plate separation $d$ and plate area $A$ , its capacitance becomes. For metal sheet,
$t = \frac{d}{2}, k = \infty$
$∴ C_{2} = \frac{ε_{0} A}{d - t}$
$= \frac{ε_{0} A}{d - \frac{d}{2}} [∵ t = \frac{d}{2}]$
$C_{2} = 2 \frac{ε_{0} A}{d}$
$C_{2} = 2 C$

[Karnataka CET-2001]

Capacitance

165627 $n$ identical capacitors each of capacitance $C$ when connected in parallel give the effective capacitance $90 μ F$ and when connected in series give $2.5 μ F$ . Then the values of $n$ and $C$ respectively are

1 6 and

15 μ F

2 5 and

18 μ F

3 15 and

6 μ F

4 18 and

5 μ F

Explanation:

:Given that,
When connected in parallel
${(C_{eq})}_{1} = 90 μ F$
' $n$ ' identical capacitance in parallel,
$(C_{eq}) = C_{1} + C_{2} + \dots \dots . . C_{n}$
$(C_{eq}) = nC \Rightarrow 90 = nC$
$n = \frac{90}{C}$
When ' $n$ ' identical capacitor connected in series,
$\frac{1}{C_{e q}} = \frac{n}{C}$
$C_{eq} = \frac{C}{n}$
$2.5 μ F = \frac{C}{n}$
$C = 2.5 \times n$
$C = 2.5 \times \frac{90}{C} [from equation(ii)]$
$C^{2} = 225$
$C = 15 μ F$
$C = 15 μ F in equation (i) we get -$
$n \times 15 = 90$
$n = 6$
Putting $C = 15 μ F$ in equation (i) we get -

[J and K CET- 2010]

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: