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Capacitance

165843 In the circuit given below, the charge in $μ C$ , on the capacitor having $5 μ F$ is

1 4.5

2 9

3 7

4 15

Explanation:

:

$6 V$
Here, $2 μ F$ and $5 μ F$ are in parallel,
Then, $C_{net} = 5 + 2 = 7 μ F$

We know that,
$Q = C \times V$
$V = \frac{Q}{C}$
$V \propto \frac{1}{C}$
Then, $V_{1} : V_{2} = C_{2} : C_{1}$
$V_{2} = \frac{C_{1}}{C_{1} + C_{2}} \times V_{1}$
$V_{2} = \frac{3}{10} \times 6$
$V_{2} = 1.8 Volt$
Hence, charge on $(Q_{5 μ F}) = C \times V$
$= 5 \times 1.8 = 9 μ C$

[CG PET- 2007

Capacitance

165844 In the given figure, the charge on $3 μ F$ capacitor is

1

10 μ C

2

15 μ C

3

30 μ C

4

5 μ C

Explanation:

:

According to Figure, all the capacitors are connected in series,
Hence, net capacitance is,
$C_{net} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\frac{1}{C} = \frac{3 + 2 + 1}{6} = \frac{6}{6}$
$C_{net} = 1 μ F$
Being series each capacitor is same charge,
Hence,
$Q = C_{net} \cdot V (V = 10 V)$
$= 1 \times 10$
$Q = 10 μ$

[BITSAT-2019]

Capacitance

165845 In the given circuit the potential drop across the capacitor must be

1

V

2

\frac{V}{2}

3

\frac{V}{3}

4

\frac{2 V}{3}

Explanation:

: Given,

In steady state, no current will pass through the capacitor,
In the lower loop,
$2 V - 2 iR - iR - V = 0$
$i = V / 3 R$
For the upper loop,
$V - V_{C} - i R - V = 0$
$V_{C} = i R$
$| V_{C} | = \frac{V}{3 R} \times R$
$| V_{C} | = \frac{V}{3}$

[BITSAT-2016]

Capacitance

165846 In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state

1

V_{AB} = V_{BC} = 100 V

2

V_{AB} = 75 V, V_{BC} = 25 V

3

V_{AB} = 25 V, V_{BC} = 75 V

4

V_{AB} = V_{BC} = 50 V

Explanation:

: Given,

At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected.
When capacitors are connected in series with a battery of potential V.
Potential difference across each capacitor is
$V_{1} = [\frac{C_{2}}{C_{1} + C_{2}}] V$
$V_{2} = [\frac{C_{1}}{C_{1} + C_{2}}] V$
Now,
$V_{1} = V_{AB} = [\frac{2}{2 + 6}] \times 100 = 25 V$
$V_{2} = V_{BC} = [\frac{6}{2 + 6}] \times 100 = 75 V$
So, correct answer is option (c).

[BITSAT-2016]

Capacitance

165843 In the circuit given below, the charge in $μ C$ , on the capacitor having $5 μ F$ is

1 4.5

2 9

3 7

4 15

Explanation:

:

$6 V$
Here, $2 μ F$ and $5 μ F$ are in parallel,
Then, $C_{net} = 5 + 2 = 7 μ F$

We know that,
$Q = C \times V$
$V = \frac{Q}{C}$
$V \propto \frac{1}{C}$
Then, $V_{1} : V_{2} = C_{2} : C_{1}$
$V_{2} = \frac{C_{1}}{C_{1} + C_{2}} \times V_{1}$
$V_{2} = \frac{3}{10} \times 6$
$V_{2} = 1.8 Volt$
Hence, charge on $(Q_{5 μ F}) = C \times V$
$= 5 \times 1.8 = 9 μ C$

[CG PET- 2007

Capacitance

165844 In the given figure, the charge on $3 μ F$ capacitor is

1

10 μ C

2

15 μ C

3

30 μ C

4

5 μ C

Explanation:

:

According to Figure, all the capacitors are connected in series,
Hence, net capacitance is,
$C_{net} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\frac{1}{C} = \frac{3 + 2 + 1}{6} = \frac{6}{6}$
$C_{net} = 1 μ F$
Being series each capacitor is same charge,
Hence,
$Q = C_{net} \cdot V (V = 10 V)$
$= 1 \times 10$
$Q = 10 μ$

[BITSAT-2019]

Capacitance

165845 In the given circuit the potential drop across the capacitor must be

1

V

2

\frac{V}{2}

3

\frac{V}{3}

4

\frac{2 V}{3}

Explanation:

: Given,

In steady state, no current will pass through the capacitor,
In the lower loop,
$2 V - 2 iR - iR - V = 0$
$i = V / 3 R$
For the upper loop,
$V - V_{C} - i R - V = 0$
$V_{C} = i R$
$| V_{C} | = \frac{V}{3 R} \times R$
$| V_{C} | = \frac{V}{3}$

[BITSAT-2016]

Capacitance

165846 In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state

1

V_{AB} = V_{BC} = 100 V

2

V_{AB} = 75 V, V_{BC} = 25 V

3

V_{AB} = 25 V, V_{BC} = 75 V

4

V_{AB} = V_{BC} = 50 V

Explanation:

: Given,

At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected.
When capacitors are connected in series with a battery of potential V.
Potential difference across each capacitor is
$V_{1} = [\frac{C_{2}}{C_{1} + C_{2}}] V$
$V_{2} = [\frac{C_{1}}{C_{1} + C_{2}}] V$
Now,
$V_{1} = V_{AB} = [\frac{2}{2 + 6}] \times 100 = 25 V$
$V_{2} = V_{BC} = [\frac{6}{2 + 6}] \times 100 = 75 V$
So, correct answer is option (c).

[BITSAT-2016]

Capacitance

165843 In the circuit given below, the charge in $μ C$ , on the capacitor having $5 μ F$ is

1 4.5

2 9

3 7

4 15

Explanation:

:

$6 V$
Here, $2 μ F$ and $5 μ F$ are in parallel,
Then, $C_{net} = 5 + 2 = 7 μ F$

We know that,
$Q = C \times V$
$V = \frac{Q}{C}$
$V \propto \frac{1}{C}$
Then, $V_{1} : V_{2} = C_{2} : C_{1}$
$V_{2} = \frac{C_{1}}{C_{1} + C_{2}} \times V_{1}$
$V_{2} = \frac{3}{10} \times 6$
$V_{2} = 1.8 Volt$
Hence, charge on $(Q_{5 μ F}) = C \times V$
$= 5 \times 1.8 = 9 μ C$

[CG PET- 2007

Capacitance

165844 In the given figure, the charge on $3 μ F$ capacitor is

1

10 μ C

2

15 μ C

3

30 μ C

4

5 μ C

Explanation:

:

According to Figure, all the capacitors are connected in series,
Hence, net capacitance is,
$C_{net} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\frac{1}{C} = \frac{3 + 2 + 1}{6} = \frac{6}{6}$
$C_{net} = 1 μ F$
Being series each capacitor is same charge,
Hence,
$Q = C_{net} \cdot V (V = 10 V)$
$= 1 \times 10$
$Q = 10 μ$

[BITSAT-2019]

Capacitance

165845 In the given circuit the potential drop across the capacitor must be

1

V

2

\frac{V}{2}

3

\frac{V}{3}

4

\frac{2 V}{3}

Explanation:

: Given,

In steady state, no current will pass through the capacitor,
In the lower loop,
$2 V - 2 iR - iR - V = 0$
$i = V / 3 R$
For the upper loop,
$V - V_{C} - i R - V = 0$
$V_{C} = i R$
$| V_{C} | = \frac{V}{3 R} \times R$
$| V_{C} | = \frac{V}{3}$

[BITSAT-2016]

Capacitance

165846 In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state

1

V_{AB} = V_{BC} = 100 V

2

V_{AB} = 75 V, V_{BC} = 25 V

3

V_{AB} = 25 V, V_{BC} = 75 V

4

V_{AB} = V_{BC} = 50 V

Explanation:

: Given,

At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected.
When capacitors are connected in series with a battery of potential V.
Potential difference across each capacitor is
$V_{1} = [\frac{C_{2}}{C_{1} + C_{2}}] V$
$V_{2} = [\frac{C_{1}}{C_{1} + C_{2}}] V$
Now,
$V_{1} = V_{AB} = [\frac{2}{2 + 6}] \times 100 = 25 V$
$V_{2} = V_{BC} = [\frac{6}{2 + 6}] \times 100 = 75 V$
So, correct answer is option (c).

[BITSAT-2016]

Capacitance

165843 In the circuit given below, the charge in $μ C$ , on the capacitor having $5 μ F$ is

1 4.5

2 9

3 7

4 15

Explanation:

:

$6 V$
Here, $2 μ F$ and $5 μ F$ are in parallel,
Then, $C_{net} = 5 + 2 = 7 μ F$

We know that,
$Q = C \times V$
$V = \frac{Q}{C}$
$V \propto \frac{1}{C}$
Then, $V_{1} : V_{2} = C_{2} : C_{1}$
$V_{2} = \frac{C_{1}}{C_{1} + C_{2}} \times V_{1}$
$V_{2} = \frac{3}{10} \times 6$
$V_{2} = 1.8 Volt$
Hence, charge on $(Q_{5 μ F}) = C \times V$
$= 5 \times 1.8 = 9 μ C$

[CG PET- 2007

Capacitance

165844 In the given figure, the charge on $3 μ F$ capacitor is

1

10 μ C

2

15 μ C

3

30 μ C

4

5 μ C

Explanation:

:

According to Figure, all the capacitors are connected in series,
Hence, net capacitance is,
$C_{net} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$
$\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\frac{1}{C} = \frac{3 + 2 + 1}{6} = \frac{6}{6}$
$C_{net} = 1 μ F$
Being series each capacitor is same charge,
Hence,
$Q = C_{net} \cdot V (V = 10 V)$
$= 1 \times 10$
$Q = 10 μ$

[BITSAT-2019]

Capacitance

165845 In the given circuit the potential drop across the capacitor must be

1

V

2

\frac{V}{2}

3

\frac{V}{3}

4

\frac{2 V}{3}

Explanation:

: Given,

In steady state, no current will pass through the capacitor,
In the lower loop,
$2 V - 2 iR - iR - V = 0$
$i = V / 3 R$
For the upper loop,
$V - V_{C} - i R - V = 0$
$V_{C} = i R$
$| V_{C} | = \frac{V}{3 R} \times R$
$| V_{C} | = \frac{V}{3}$

[BITSAT-2016]

Capacitance

165846 In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state

1

V_{AB} = V_{BC} = 100 V

2

V_{AB} = 75 V, V_{BC} = 25 V

3

V_{AB} = 25 V, V_{BC} = 75 V

4

V_{AB} = V_{BC} = 50 V

Explanation:

: Given,

At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected.
When capacitors are connected in series with a battery of potential V.
Potential difference across each capacitor is
$V_{1} = [\frac{C_{2}}{C_{1} + C_{2}}] V$
$V_{2} = [\frac{C_{1}}{C_{1} + C_{2}}] V$
Now,
$V_{1} = V_{AB} = [\frac{2}{2 + 6}] \times 100 = 25 V$
$V_{2} = V_{BC} = [\frac{6}{2 + 6}] \times 100 = 75 V$
So, correct answer is option (c).

[BITSAT-2016]