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PHXII03:CURRENT ELECTRICITY

357477 supporting img
In the given circuit, find charge on capacitor after $1 s$ of opening the switch at $t = \infty$ .

1

20 e^{- 10} μ C

2

25 e^{- 10} μ C

3

30 e^{- 10} μ C

4

35 e^{- 10} μ C

Explanation:

At $t \to \infty$ , capacitor
Works as on open circuit as shown in the figure.
supporting img
Therefore, current flow in the circuit,
$I_{1} = \frac{9}{(12 + 15) 10^{3}} = \frac{1}{3} \times 10^{- 3} A$
$∴ V_{0} = I_{1} \times 15 \times 10^{3}$
$= \frac{1}{3} \times 10^{- 3} \times 15 \times 10^{3} = 5 V$
$∴$ Charge on capacitor of $5 μ F$ , $q_{0} = C V_{0} = 5 \times 10^{- 6} \times 5$
$= 25 \times 10^{- 6} C = 25 μ C$
When switch is open, then capacitor starts discharging across
$(5 + 15 = 20 k Ω)$ and instantaneous charge on capacitor is given by
$q = q_{0} e^{- t / R C}$
$= 25 e^{- \frac{1}{20 \times 10^{3} \times 5 \times 10^{- 6}}} [∵ t = 1 s]$
$= 25 e^{- 10} μ C$

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357478 The capacitance of the system is $C$ ; if the key is closed, the total energy loss is equal to:
supporting img

1

\frac{Q^{2}}{2 C}

2

\frac{2 Q^{2}}{C}

3

\frac{9 Q^{2}}{8 C}

4 None of these

Explanation:

The charge distribution on the plates is
Energy stored between the plates
supporting img
$= \frac{{(3 Q / 2)}^{2}}{2 C} = \frac{9 Q^{2}}{8 C}$
The entire energy which is stored between the plates is lost after closing the key- $K$

PHXII03:CURRENT ELECTRICITY

357479 In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
supporting img

1

13 μ C

2

12 μ C

3

9 μ C

4

Z e r o

Explanation:

Sum of charges on plates of capacitor was zero initially and will be zero finally
$∴$ No charge flows to the ground.

PHXII03:CURRENT ELECTRICITY

357480 A galvanometer $(G)$ of $2 Ω$ resistance is connected in the given circuit. The ratio of charges stored in $C_{1}$ and $C_{2}$ is
supporting img

1 1

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

In steady state, capacitor behaves like an open circuit.
So, $R_{e q} = 4 + 6 + 2 = 12 Ω$
Current, $I = \frac{6}{12} \Rightarrow 0.5 A$
Potential difference across $C_{1} =$ Potential difference across $4 Ω$
supporting img

$V_{1} = I (4 + R_{G})$
$∴ V_{1} = 0.5 \times 6 = 3 V$
Charge stored in $C_{1}, Q_{1} = C_{1} V_{1}$
So, $Q_{1} = 4 \times 10^{- 6} \times 3 = 12 \times 10^{- 6} = 12 μ C$
Potential difference across $C_{2} =$ Potential difference across $6 Ω$
$V_{2} = I (6 + R_{G})$
$∴ V_{2} = 0.5 \times 8 = 4 V$
Charge stored in $C_{2}, Q_{2} = C_{2} V_{2}$
$Q_{2} = 6 \times 10^{- 6} \times 4 = 24 μ C$
supporting img

$∴ Q_{1} : Q_{2} = 12 : 24 = 1 : 2$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357481 A capacitor of 8 $F$ is connected as shown. Charge on the plates of the capacitor
supporting img

1

32 C

2

40 C

3

0 C

4

80 C

Explanation:

At steady state, the capacitor gets fully charged and stops conducting.
$∴$ Current through the $20 Ω$ resistance is zero
$∴$ Current in the circuit,
$I = \frac{5}{4 + 1} = 1 A$
Voltage across the $4 Ω$ resistance
$= V = I \times 4 = 4 V$
= Voltage across the capacitor.
Hence charge on capacitor,
$q = C V = 8 \times 4 = 32 C$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

357477 supporting img
In the given circuit, find charge on capacitor after $1 s$ of opening the switch at $t = \infty$ .

1

20 e^{- 10} μ C

2

25 e^{- 10} μ C

3

30 e^{- 10} μ C

4

35 e^{- 10} μ C

Explanation:

At $t \to \infty$ , capacitor
Works as on open circuit as shown in the figure.
supporting img
Therefore, current flow in the circuit,
$I_{1} = \frac{9}{(12 + 15) 10^{3}} = \frac{1}{3} \times 10^{- 3} A$
$∴ V_{0} = I_{1} \times 15 \times 10^{3}$
$= \frac{1}{3} \times 10^{- 3} \times 15 \times 10^{3} = 5 V$
$∴$ Charge on capacitor of $5 μ F$ , $q_{0} = C V_{0} = 5 \times 10^{- 6} \times 5$
$= 25 \times 10^{- 6} C = 25 μ C$
When switch is open, then capacitor starts discharging across
$(5 + 15 = 20 k Ω)$ and instantaneous charge on capacitor is given by
$q = q_{0} e^{- t / R C}$
$= 25 e^{- \frac{1}{20 \times 10^{3} \times 5 \times 10^{- 6}}} [∵ t = 1 s]$
$= 25 e^{- 10} μ C$

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357478 The capacitance of the system is $C$ ; if the key is closed, the total energy loss is equal to:
supporting img

1

\frac{Q^{2}}{2 C}

2

\frac{2 Q^{2}}{C}

3

\frac{9 Q^{2}}{8 C}

4 None of these

Explanation:

The charge distribution on the plates is
Energy stored between the plates
supporting img
$= \frac{{(3 Q / 2)}^{2}}{2 C} = \frac{9 Q^{2}}{8 C}$
The entire energy which is stored between the plates is lost after closing the key- $K$

PHXII03:CURRENT ELECTRICITY

357479 In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
supporting img

1

13 μ C

2

12 μ C

3

9 μ C

4

Z e r o

Explanation:

Sum of charges on plates of capacitor was zero initially and will be zero finally
$∴$ No charge flows to the ground.

PHXII03:CURRENT ELECTRICITY

357480 A galvanometer $(G)$ of $2 Ω$ resistance is connected in the given circuit. The ratio of charges stored in $C_{1}$ and $C_{2}$ is
supporting img

1 1

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

In steady state, capacitor behaves like an open circuit.
So, $R_{e q} = 4 + 6 + 2 = 12 Ω$
Current, $I = \frac{6}{12} \Rightarrow 0.5 A$
Potential difference across $C_{1} =$ Potential difference across $4 Ω$
supporting img

$V_{1} = I (4 + R_{G})$
$∴ V_{1} = 0.5 \times 6 = 3 V$
Charge stored in $C_{1}, Q_{1} = C_{1} V_{1}$
So, $Q_{1} = 4 \times 10^{- 6} \times 3 = 12 \times 10^{- 6} = 12 μ C$
Potential difference across $C_{2} =$ Potential difference across $6 Ω$
$V_{2} = I (6 + R_{G})$
$∴ V_{2} = 0.5 \times 8 = 4 V$
Charge stored in $C_{2}, Q_{2} = C_{2} V_{2}$
$Q_{2} = 6 \times 10^{- 6} \times 4 = 24 μ C$
supporting img

$∴ Q_{1} : Q_{2} = 12 : 24 = 1 : 2$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357481 A capacitor of 8 $F$ is connected as shown. Charge on the plates of the capacitor
supporting img

1

32 C

2

40 C

3

0 C

4

80 C

Explanation:

At steady state, the capacitor gets fully charged and stops conducting.
$∴$ Current through the $20 Ω$ resistance is zero
$∴$ Current in the circuit,
$I = \frac{5}{4 + 1} = 1 A$
Voltage across the $4 Ω$ resistance
$= V = I \times 4 = 4 V$
= Voltage across the capacitor.
Hence charge on capacitor,
$q = C V = 8 \times 4 = 32 C$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

357477 supporting img
In the given circuit, find charge on capacitor after $1 s$ of opening the switch at $t = \infty$ .

1

20 e^{- 10} μ C

2

25 e^{- 10} μ C

3

30 e^{- 10} μ C

4

35 e^{- 10} μ C

Explanation:

At $t \to \infty$ , capacitor
Works as on open circuit as shown in the figure.
supporting img
Therefore, current flow in the circuit,
$I_{1} = \frac{9}{(12 + 15) 10^{3}} = \frac{1}{3} \times 10^{- 3} A$
$∴ V_{0} = I_{1} \times 15 \times 10^{3}$
$= \frac{1}{3} \times 10^{- 3} \times 15 \times 10^{3} = 5 V$
$∴$ Charge on capacitor of $5 μ F$ , $q_{0} = C V_{0} = 5 \times 10^{- 6} \times 5$
$= 25 \times 10^{- 6} C = 25 μ C$
When switch is open, then capacitor starts discharging across
$(5 + 15 = 20 k Ω)$ and instantaneous charge on capacitor is given by
$q = q_{0} e^{- t / R C}$
$= 25 e^{- \frac{1}{20 \times 10^{3} \times 5 \times 10^{- 6}}} [∵ t = 1 s]$
$= 25 e^{- 10} μ C$

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357478 The capacitance of the system is $C$ ; if the key is closed, the total energy loss is equal to:
supporting img

1

\frac{Q^{2}}{2 C}

2

\frac{2 Q^{2}}{C}

3

\frac{9 Q^{2}}{8 C}

4 None of these

Explanation:

The charge distribution on the plates is
Energy stored between the plates
supporting img
$= \frac{{(3 Q / 2)}^{2}}{2 C} = \frac{9 Q^{2}}{8 C}$
The entire energy which is stored between the plates is lost after closing the key- $K$

PHXII03:CURRENT ELECTRICITY

357479 In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
supporting img

1

13 μ C

2

12 μ C

3

9 μ C

4

Z e r o

Explanation:

Sum of charges on plates of capacitor was zero initially and will be zero finally
$∴$ No charge flows to the ground.

PHXII03:CURRENT ELECTRICITY

357480 A galvanometer $(G)$ of $2 Ω$ resistance is connected in the given circuit. The ratio of charges stored in $C_{1}$ and $C_{2}$ is
supporting img

1 1

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

In steady state, capacitor behaves like an open circuit.
So, $R_{e q} = 4 + 6 + 2 = 12 Ω$
Current, $I = \frac{6}{12} \Rightarrow 0.5 A$
Potential difference across $C_{1} =$ Potential difference across $4 Ω$
supporting img

$V_{1} = I (4 + R_{G})$
$∴ V_{1} = 0.5 \times 6 = 3 V$
Charge stored in $C_{1}, Q_{1} = C_{1} V_{1}$
So, $Q_{1} = 4 \times 10^{- 6} \times 3 = 12 \times 10^{- 6} = 12 μ C$
Potential difference across $C_{2} =$ Potential difference across $6 Ω$
$V_{2} = I (6 + R_{G})$
$∴ V_{2} = 0.5 \times 8 = 4 V$
Charge stored in $C_{2}, Q_{2} = C_{2} V_{2}$
$Q_{2} = 6 \times 10^{- 6} \times 4 = 24 μ C$
supporting img

$∴ Q_{1} : Q_{2} = 12 : 24 = 1 : 2$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357481 A capacitor of 8 $F$ is connected as shown. Charge on the plates of the capacitor
supporting img

1

32 C

2

40 C

3

0 C

4

80 C

Explanation:

At steady state, the capacitor gets fully charged and stops conducting.
$∴$ Current through the $20 Ω$ resistance is zero
$∴$ Current in the circuit,
$I = \frac{5}{4 + 1} = 1 A$
Voltage across the $4 Ω$ resistance
$= V = I \times 4 = 4 V$
= Voltage across the capacitor.
Hence charge on capacitor,
$q = C V = 8 \times 4 = 32 C$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

357477 supporting img
In the given circuit, find charge on capacitor after $1 s$ of opening the switch at $t = \infty$ .

1

20 e^{- 10} μ C

2

25 e^{- 10} μ C

3

30 e^{- 10} μ C

4

35 e^{- 10} μ C

Explanation:

At $t \to \infty$ , capacitor
Works as on open circuit as shown in the figure.
supporting img
Therefore, current flow in the circuit,
$I_{1} = \frac{9}{(12 + 15) 10^{3}} = \frac{1}{3} \times 10^{- 3} A$
$∴ V_{0} = I_{1} \times 15 \times 10^{3}$
$= \frac{1}{3} \times 10^{- 3} \times 15 \times 10^{3} = 5 V$
$∴$ Charge on capacitor of $5 μ F$ , $q_{0} = C V_{0} = 5 \times 10^{- 6} \times 5$
$= 25 \times 10^{- 6} C = 25 μ C$
When switch is open, then capacitor starts discharging across
$(5 + 15 = 20 k Ω)$ and instantaneous charge on capacitor is given by
$q = q_{0} e^{- t / R C}$
$= 25 e^{- \frac{1}{20 \times 10^{3} \times 5 \times 10^{- 6}}} [∵ t = 1 s]$
$= 25 e^{- 10} μ C$

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357478 The capacitance of the system is $C$ ; if the key is closed, the total energy loss is equal to:
supporting img

1

\frac{Q^{2}}{2 C}

2

\frac{2 Q^{2}}{C}

3

\frac{9 Q^{2}}{8 C}

4 None of these

Explanation:

The charge distribution on the plates is
Energy stored between the plates
supporting img
$= \frac{{(3 Q / 2)}^{2}}{2 C} = \frac{9 Q^{2}}{8 C}$
The entire energy which is stored between the plates is lost after closing the key- $K$

PHXII03:CURRENT ELECTRICITY

357479 In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
supporting img

1

13 μ C

2

12 μ C

3

9 μ C

4

Z e r o

Explanation:

Sum of charges on plates of capacitor was zero initially and will be zero finally
$∴$ No charge flows to the ground.

PHXII03:CURRENT ELECTRICITY

357480 A galvanometer $(G)$ of $2 Ω$ resistance is connected in the given circuit. The ratio of charges stored in $C_{1}$ and $C_{2}$ is
supporting img

1 1

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

In steady state, capacitor behaves like an open circuit.
So, $R_{e q} = 4 + 6 + 2 = 12 Ω$
Current, $I = \frac{6}{12} \Rightarrow 0.5 A$
Potential difference across $C_{1} =$ Potential difference across $4 Ω$
supporting img

$V_{1} = I (4 + R_{G})$
$∴ V_{1} = 0.5 \times 6 = 3 V$
Charge stored in $C_{1}, Q_{1} = C_{1} V_{1}$
So, $Q_{1} = 4 \times 10^{- 6} \times 3 = 12 \times 10^{- 6} = 12 μ C$
Potential difference across $C_{2} =$ Potential difference across $6 Ω$
$V_{2} = I (6 + R_{G})$
$∴ V_{2} = 0.5 \times 8 = 4 V$
Charge stored in $C_{2}, Q_{2} = C_{2} V_{2}$
$Q_{2} = 6 \times 10^{- 6} \times 4 = 24 μ C$
supporting img

$∴ Q_{1} : Q_{2} = 12 : 24 = 1 : 2$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357481 A capacitor of 8 $F$ is connected as shown. Charge on the plates of the capacitor
supporting img

1

32 C

2

40 C

3

0 C

4

80 C

Explanation:

At steady state, the capacitor gets fully charged and stops conducting.
$∴$ Current through the $20 Ω$ resistance is zero
$∴$ Current in the circuit,
$I = \frac{5}{4 + 1} = 1 A$
Voltage across the $4 Ω$ resistance
$= V = I \times 4 = 4 V$
= Voltage across the capacitor.
Hence charge on capacitor,
$q = C V = 8 \times 4 = 32 C$
supporting img

KCET - 2016

PHXII03:CURRENT ELECTRICITY

357477 supporting img
In the given circuit, find charge on capacitor after $1 s$ of opening the switch at $t = \infty$ .

1

20 e^{- 10} μ C

2

25 e^{- 10} μ C

3

30 e^{- 10} μ C

4

35 e^{- 10} μ C

Explanation:

At $t \to \infty$ , capacitor
Works as on open circuit as shown in the figure.
supporting img
Therefore, current flow in the circuit,
$I_{1} = \frac{9}{(12 + 15) 10^{3}} = \frac{1}{3} \times 10^{- 3} A$
$∴ V_{0} = I_{1} \times 15 \times 10^{3}$
$= \frac{1}{3} \times 10^{- 3} \times 15 \times 10^{3} = 5 V$
$∴$ Charge on capacitor of $5 μ F$ , $q_{0} = C V_{0} = 5 \times 10^{- 6} \times 5$
$= 25 \times 10^{- 6} C = 25 μ C$
When switch is open, then capacitor starts discharging across
$(5 + 15 = 20 k Ω)$ and instantaneous charge on capacitor is given by
$q = q_{0} e^{- t / R C}$
$= 25 e^{- \frac{1}{20 \times 10^{3} \times 5 \times 10^{- 6}}} [∵ t = 1 s]$
$= 25 e^{- 10} μ C$

AIIMS - 2019

PHXII03:CURRENT ELECTRICITY

357478 The capacitance of the system is $C$ ; if the key is closed, the total energy loss is equal to:
supporting img

1

\frac{Q^{2}}{2 C}

2

\frac{2 Q^{2}}{C}

3

\frac{9 Q^{2}}{8 C}

4 None of these

Explanation:

The charge distribution on the plates is
Energy stored between the plates
supporting img
$= \frac{{(3 Q / 2)}^{2}}{2 C} = \frac{9 Q^{2}}{8 C}$
The entire energy which is stored between the plates is lost after closing the key- $K$

PHXII03:CURRENT ELECTRICITY

357479 In the given circuit, the quantity of charge that flows to ground long time after the switch is closed is:
supporting img

1

13 μ C

2

12 μ C

3

9 μ C

4

Z e r o

Explanation:

Sum of charges on plates of capacitor was zero initially and will be zero finally
$∴$ No charge flows to the ground.

PHXII03:CURRENT ELECTRICITY

357480 A galvanometer $(G)$ of $2 Ω$ resistance is connected in the given circuit. The ratio of charges stored in $C_{1}$ and $C_{2}$ is
supporting img

1 1

2

\frac{2}{3}

3

\frac{3}{2}

4

\frac{1}{2}

Explanation:

In steady state, capacitor behaves like an open circuit.
So, $R_{e q} = 4 + 6 + 2 = 12 Ω$
Current, $I = \frac{6}{12} \Rightarrow 0.5 A$
Potential difference across $C_{1} =$ Potential difference across $4 Ω$
supporting img

$V_{1} = I (4 + R_{G})$
$∴ V_{1} = 0.5 \times 6 = 3 V$
Charge stored in $C_{1}, Q_{1} = C_{1} V_{1}$
So, $Q_{1} = 4 \times 10^{- 6} \times 3 = 12 \times 10^{- 6} = 12 μ C$
Potential difference across $C_{2} =$ Potential difference across $6 Ω$
$V_{2} = I (6 + R_{G})$
$∴ V_{2} = 0.5 \times 8 = 4 V$
Charge stored in $C_{2}, Q_{2} = C_{2} V_{2}$
$Q_{2} = 6 \times 10^{- 6} \times 4 = 24 μ C$
supporting img

$∴ Q_{1} : Q_{2} = 12 : 24 = 1 : 2$

JEE - 2024

PHXII03:CURRENT ELECTRICITY

357481 A capacitor of 8 $F$ is connected as shown. Charge on the plates of the capacitor
supporting img

1

32 C

2

40 C

3

0 C

4

80 C

Explanation:

At steady state, the capacitor gets fully charged and stops conducting.
$∴$ Current through the $20 Ω$ resistance is zero
$∴$ Current in the circuit,
$I = \frac{5}{4 + 1} = 1 A$
Voltage across the $4 Ω$ resistance
$= V = I \times 4 = 4 V$
= Voltage across the capacitor.
Hence charge on capacitor,
$q = C V = 8 \times 4 = 32 C$
supporting img

KCET - 2016

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