QBManager

Capacitance

165814 In the given circuit, if the potential difference between $A$ and $B$ is $80 V$ , then the equivalent capacitance between $A$ and $B$ , and the charge on $10 μ F$ capacitor respectively, are

1

4 μ F & 133 μ C

2

164 μ F & 150 μ C

3

15 μ F & 200 μ C

4

4 μ F & 50 μ C

Explanation:

: From given figure,
Redrawn the circuit -

Since, $12 μ F, 24 μ F$ and $8 μ F$ are in series.
So, $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8}$
$\frac{1}{C_{eq}} = \frac{2 + 1 + 3}{24}$
$\frac{1}{C_{eq}} = \frac{6}{24} = \frac{1}{4}$
$C_{eq} = 4 μ F$
Charge, $Q = C_{eq} \times V$
$= 4 \times 80 \times 10^{- 6} = 320 μ C$
Charge in $10 μ F$ capacitor,
$Q^{'} = \frac{10}{10 + 5 + 9} \times 320$
$= \frac{10}{24} \times 320$
$= 133.33 μ C .$
$◻ 133 μ C$

[AP EAMCET-19.08.2021

Capacitance

165815 Three capacitors of capacitances $C_{1} = 2 μ F, C_{2} =$ $3 μ F$ and $C_{3} = 5 μ F$ are connected in series. $A$ potential difference of $155 V$ is applied across the combination. Choose the correct option.

1 Least potential difference is across

C_{3}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

75 V

.

2 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{51}) μ F

and the voltage across

C_{2}

is

50 V

.

3 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{3}

is

30 V

.

4 Least potential difference is across

C_{2}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

50 V

.

Explanation:

:

Given
$C_{1} = 2 μ F, C_{2} = 3 μ F, C_{3} = 5 μ F$
Equivalent capacitance in series combination given by:-
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$C_{eq} = \frac{30}{31} μ F$
$Q = C_{eq} \times V$
$Q = \frac{30}{31} \times 155 \times 10^{- 6}$
$Q = 150 μ C$
So, voltage across $C_{1}$ is
$V_{1} = \frac{Q}{C_{1}}$
$V_{1} = \frac{150}{2} = 75 V$
Therefore, Least potential difference is across $C_{3}$ .
Equivalent capacitance of combination is $(\frac{30}{31}) μ F$ and the voltage across $C_{1}$ is $75 V$ .

[AP EAMCET (17.09.2020) Shift-II]

Capacitance

165817 Assertion: If three capacitors of capacitances $C_{1} <$ $C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $Cp > Cs$ .
Reason: $\frac{1}{C_{p}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion. (c) If both the Assertion is correct but Reason is incorrect.

3 (d.) If both the Assertion and Reason are incorrect.

4 If the Assertion is incorrect but the Reason is correct.

Explanation:

: In a parallel plate, capacitance will increase.
$C_{parallel} = C_{1} + C_{2} + C_{3}$
In a series, capacitance will be less than any parallel plate capacitor
$C_{series} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
So, $C_{parallel} > C_{series}$
Hence, assertion is correct and reason is incorrect.

[AIIMS-2002]

Capacitance

165818 For the circuit shown in the figure the charge on $2 μ F$ capacitor is

1

15 μ C

2

20 μ C

3

30 μ C

4

45 μ C

Explanation:

:

$Q = q_{1} + q_{2}$
$2 V_{0} = 1 (10 - V_{0}) + 5 (10 - V_{0})$
$2 V_{0} = 10 - V_{0} + 50 - 5 V_{0}$
$8 V_{0} = 60$
$V_{0} = \frac{60}{8}$
$∵ Q = {CV}_{o}$
$Q = 2 V_{0} = 2 \times \frac{60}{8}$
$∴ Q = \frac{60}{4}$
$Q = 15 μ C$

[AIIMS-27.05.2018(E)]

Capacitance

165814 In the given circuit, if the potential difference between $A$ and $B$ is $80 V$ , then the equivalent capacitance between $A$ and $B$ , and the charge on $10 μ F$ capacitor respectively, are

1

4 μ F & 133 μ C

2

164 μ F & 150 μ C

3

15 μ F & 200 μ C

4

4 μ F & 50 μ C

Explanation:

: From given figure,
Redrawn the circuit -

Since, $12 μ F, 24 μ F$ and $8 μ F$ are in series.
So, $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8}$
$\frac{1}{C_{eq}} = \frac{2 + 1 + 3}{24}$
$\frac{1}{C_{eq}} = \frac{6}{24} = \frac{1}{4}$
$C_{eq} = 4 μ F$
Charge, $Q = C_{eq} \times V$
$= 4 \times 80 \times 10^{- 6} = 320 μ C$
Charge in $10 μ F$ capacitor,
$Q^{'} = \frac{10}{10 + 5 + 9} \times 320$
$= \frac{10}{24} \times 320$
$= 133.33 μ C .$
$◻ 133 μ C$

[AP EAMCET-19.08.2021

Capacitance

165815 Three capacitors of capacitances $C_{1} = 2 μ F, C_{2} =$ $3 μ F$ and $C_{3} = 5 μ F$ are connected in series. $A$ potential difference of $155 V$ is applied across the combination. Choose the correct option.

1 Least potential difference is across

C_{3}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

75 V

.

2 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{51}) μ F

and the voltage across

C_{2}

is

50 V

.

3 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{3}

is

30 V

.

4 Least potential difference is across

C_{2}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

50 V

.

Explanation:

:

Given
$C_{1} = 2 μ F, C_{2} = 3 μ F, C_{3} = 5 μ F$
Equivalent capacitance in series combination given by:-
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$C_{eq} = \frac{30}{31} μ F$
$Q = C_{eq} \times V$
$Q = \frac{30}{31} \times 155 \times 10^{- 6}$
$Q = 150 μ C$
So, voltage across $C_{1}$ is
$V_{1} = \frac{Q}{C_{1}}$
$V_{1} = \frac{150}{2} = 75 V$
Therefore, Least potential difference is across $C_{3}$ .
Equivalent capacitance of combination is $(\frac{30}{31}) μ F$ and the voltage across $C_{1}$ is $75 V$ .

[AP EAMCET (17.09.2020) Shift-II]

Capacitance

165817 Assertion: If three capacitors of capacitances $C_{1} <$ $C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $Cp > Cs$ .
Reason: $\frac{1}{C_{p}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion. (c) If both the Assertion is correct but Reason is incorrect.

3 (d.) If both the Assertion and Reason are incorrect.

4 If the Assertion is incorrect but the Reason is correct.

Explanation:

: In a parallel plate, capacitance will increase.
$C_{parallel} = C_{1} + C_{2} + C_{3}$
In a series, capacitance will be less than any parallel plate capacitor
$C_{series} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
So, $C_{parallel} > C_{series}$
Hence, assertion is correct and reason is incorrect.

[AIIMS-2002]

Capacitance

165818 For the circuit shown in the figure the charge on $2 μ F$ capacitor is

1

15 μ C

2

20 μ C

3

30 μ C

4

45 μ C

Explanation:

:

$Q = q_{1} + q_{2}$
$2 V_{0} = 1 (10 - V_{0}) + 5 (10 - V_{0})$
$2 V_{0} = 10 - V_{0} + 50 - 5 V_{0}$
$8 V_{0} = 60$
$V_{0} = \frac{60}{8}$
$∵ Q = {CV}_{o}$
$Q = 2 V_{0} = 2 \times \frac{60}{8}$
$∴ Q = \frac{60}{4}$
$Q = 15 μ C$

[AIIMS-27.05.2018(E)]

Capacitance

165814 In the given circuit, if the potential difference between $A$ and $B$ is $80 V$ , then the equivalent capacitance between $A$ and $B$ , and the charge on $10 μ F$ capacitor respectively, are

1

4 μ F & 133 μ C

2

164 μ F & 150 μ C

3

15 μ F & 200 μ C

4

4 μ F & 50 μ C

Explanation:

: From given figure,
Redrawn the circuit -

Since, $12 μ F, 24 μ F$ and $8 μ F$ are in series.
So, $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8}$
$\frac{1}{C_{eq}} = \frac{2 + 1 + 3}{24}$
$\frac{1}{C_{eq}} = \frac{6}{24} = \frac{1}{4}$
$C_{eq} = 4 μ F$
Charge, $Q = C_{eq} \times V$
$= 4 \times 80 \times 10^{- 6} = 320 μ C$
Charge in $10 μ F$ capacitor,
$Q^{'} = \frac{10}{10 + 5 + 9} \times 320$
$= \frac{10}{24} \times 320$
$= 133.33 μ C .$
$◻ 133 μ C$

[AP EAMCET-19.08.2021

Capacitance

165815 Three capacitors of capacitances $C_{1} = 2 μ F, C_{2} =$ $3 μ F$ and $C_{3} = 5 μ F$ are connected in series. $A$ potential difference of $155 V$ is applied across the combination. Choose the correct option.

1 Least potential difference is across

C_{3}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

75 V

.

2 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{51}) μ F

and the voltage across

C_{2}

is

50 V

.

3 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{3}

is

30 V

.

4 Least potential difference is across

C_{2}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

50 V

.

Explanation:

:

Given
$C_{1} = 2 μ F, C_{2} = 3 μ F, C_{3} = 5 μ F$
Equivalent capacitance in series combination given by:-
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$C_{eq} = \frac{30}{31} μ F$
$Q = C_{eq} \times V$
$Q = \frac{30}{31} \times 155 \times 10^{- 6}$
$Q = 150 μ C$
So, voltage across $C_{1}$ is
$V_{1} = \frac{Q}{C_{1}}$
$V_{1} = \frac{150}{2} = 75 V$
Therefore, Least potential difference is across $C_{3}$ .
Equivalent capacitance of combination is $(\frac{30}{31}) μ F$ and the voltage across $C_{1}$ is $75 V$ .

[AP EAMCET (17.09.2020) Shift-II]

Capacitance

165817 Assertion: If three capacitors of capacitances $C_{1} <$ $C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $Cp > Cs$ .
Reason: $\frac{1}{C_{p}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion. (c) If both the Assertion is correct but Reason is incorrect.

3 (d.) If both the Assertion and Reason are incorrect.

4 If the Assertion is incorrect but the Reason is correct.

Explanation:

: In a parallel plate, capacitance will increase.
$C_{parallel} = C_{1} + C_{2} + C_{3}$
In a series, capacitance will be less than any parallel plate capacitor
$C_{series} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
So, $C_{parallel} > C_{series}$
Hence, assertion is correct and reason is incorrect.

[AIIMS-2002]

Capacitance

165818 For the circuit shown in the figure the charge on $2 μ F$ capacitor is

1

15 μ C

2

20 μ C

3

30 μ C

4

45 μ C

Explanation:

:

$Q = q_{1} + q_{2}$
$2 V_{0} = 1 (10 - V_{0}) + 5 (10 - V_{0})$
$2 V_{0} = 10 - V_{0} + 50 - 5 V_{0}$
$8 V_{0} = 60$
$V_{0} = \frac{60}{8}$
$∵ Q = {CV}_{o}$
$Q = 2 V_{0} = 2 \times \frac{60}{8}$
$∴ Q = \frac{60}{4}$
$Q = 15 μ C$

[AIIMS-27.05.2018(E)]

Capacitance

165814 In the given circuit, if the potential difference between $A$ and $B$ is $80 V$ , then the equivalent capacitance between $A$ and $B$ , and the charge on $10 μ F$ capacitor respectively, are

1

4 μ F & 133 μ C

2

164 μ F & 150 μ C

3

15 μ F & 200 μ C

4

4 μ F & 50 μ C

Explanation:

: From given figure,
Redrawn the circuit -

Since, $12 μ F, 24 μ F$ and $8 μ F$ are in series.
So, $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8}$
$\frac{1}{C_{eq}} = \frac{2 + 1 + 3}{24}$
$\frac{1}{C_{eq}} = \frac{6}{24} = \frac{1}{4}$
$C_{eq} = 4 μ F$
Charge, $Q = C_{eq} \times V$
$= 4 \times 80 \times 10^{- 6} = 320 μ C$
Charge in $10 μ F$ capacitor,
$Q^{'} = \frac{10}{10 + 5 + 9} \times 320$
$= \frac{10}{24} \times 320$
$= 133.33 μ C .$
$◻ 133 μ C$

[AP EAMCET-19.08.2021

Capacitance

165815 Three capacitors of capacitances $C_{1} = 2 μ F, C_{2} =$ $3 μ F$ and $C_{3} = 5 μ F$ are connected in series. $A$ potential difference of $155 V$ is applied across the combination. Choose the correct option.

1 Least potential difference is across

C_{3}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

75 V

.

2 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{51}) μ F

and the voltage across

C_{2}

is

50 V

.

3 Least potential difference is across

C_{1}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{3}

is

30 V

.

4 Least potential difference is across

C_{2}

.
Equivalent capacitance of combination is

(\frac{30}{31}) μ F

and the voltage across

C_{1}

is

50 V

.

Explanation:

:

Given
$C_{1} = 2 μ F, C_{2} = 3 μ F, C_{3} = 5 μ F$
Equivalent capacitance in series combination given by:-
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
$C_{eq} = \frac{30}{31} μ F$
$Q = C_{eq} \times V$
$Q = \frac{30}{31} \times 155 \times 10^{- 6}$
$Q = 150 μ C$
So, voltage across $C_{1}$ is
$V_{1} = \frac{Q}{C_{1}}$
$V_{1} = \frac{150}{2} = 75 V$
Therefore, Least potential difference is across $C_{3}$ .
Equivalent capacitance of combination is $(\frac{30}{31}) μ F$ and the voltage across $C_{1}$ is $75 V$ .

[AP EAMCET (17.09.2020) Shift-II]

Capacitance

165817 Assertion: If three capacitors of capacitances $C_{1} <$ $C_{2} < C_{3}$ are connected in parallel then their equivalent capacitance $Cp > Cs$ .
Reason: $\frac{1}{C_{p}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but the Reason is not a correct explanation of the Assertion. (c) If both the Assertion is correct but Reason is incorrect.

3 (d.) If both the Assertion and Reason are incorrect.

4 If the Assertion is incorrect but the Reason is correct.

Explanation:

: In a parallel plate, capacitance will increase.
$C_{parallel} = C_{1} + C_{2} + C_{3}$
In a series, capacitance will be less than any parallel plate capacitor
$C_{series} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
So, $C_{parallel} > C_{series}$
Hence, assertion is correct and reason is incorrect.

[AIIMS-2002]

Capacitance

165818 For the circuit shown in the figure the charge on $2 μ F$ capacitor is

1

15 μ C

2

20 μ C

3

30 μ C

4

45 μ C

Explanation:

:

$Q = q_{1} + q_{2}$
$2 V_{0} = 1 (10 - V_{0}) + 5 (10 - V_{0})$
$2 V_{0} = 10 - V_{0} + 50 - 5 V_{0}$
$8 V_{0} = 60$
$V_{0} = \frac{60}{8}$
$∵ Q = {CV}_{o}$
$Q = 2 V_{0} = 2 \times \frac{60}{8}$
$∴ Q = \frac{60}{4}$
$Q = 15 μ C$

[AIIMS-27.05.2018(E)]