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PHXI15:WAVES

358748 A parallel plate capacitor has a capacitance $C = 200 p F .$ It is connected to $230 V$ ac supply with an angular frequency $300 r a d / s .$ The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

1

13.8 μ A

and

138 μ A

2

14.3 μ A

and

143 μ A

3

13.8 μ A

and

13.8 μ A

4

1.38 μ A

and

1.38 μ A

Explanation:

$C = 200 p F, ω = 300 r a d / s, V = 230 V$
supporting img

$I = \frac{V}{X_{C}} = ω C V = 300 \times 200 \times 10^{- 12} \times 230$
$I = 13.8 μ A$
The magnitude of conduction current and displacement current is same.

JEE - 2024

PHXI15:WAVES

358749 An alternating voltage of amplitude $40 V$ and frequency $4 k H z$ is applied directly across the capacitor of $12 μ F .$ The maximum displacement current between the plates of the capacitor is nearly

1

13 A

2

8 A

3

10 A

4

12 A

Explanation:

Given, $V_{0} = 40 V,$ $f = 4 k H z = 4 \times 10^{3} H z$ $C = 12 \times 10^{- 6} F$
Voltage at any time ' $t$ ' is given by,
$V = V_{0} \sin (ω t) = V_{0} \sin (2 π f t)$
$∴ V = 40 \sin (2 π \times 4 \times 10^{3})$
$= 40 \sin (8 π \times 10^{3} t) V$
Displacement current, $I_{d} =$ Conduction current, $I_{C}$
$∴ I_{d} = C \frac{d V}{d t}$
$= 12 \times 10^{- 6} \times [40 \times 8 π \times 10^{3} \cos (8 π \times 10^{3} t)]$
$= 12.05 \cos (8 π \times 10^{3} t)$
$∴$ Maximum displacement current,
${(I_{d})}_{max} = 12.05 A (∴ For {(I_{d})}_{max,} \cos θ = 1)$
So, option (4) is correct

JEE - 2024

PHXI15:WAVES

358750 Figure shows a capacitor made of two circular plates each of radius $12 c m$ , separated by $5.0 m m$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$ , then which of the following is correct?
(Given: $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )
supporting img

1 The rate of change of potential difference between the plates is

1.87 \times 10^{9} {V s}^{- 1}

2 The displacement current across the plates is

0.30 A

.

3 The capacitance of the capacitor is

75 p F

.

4 None of these

Explanation:

Area, $A = π R^{2} = 3.14 \times (0.12)^{2} m^{2}$
$C = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 3.14 \times (0.12)^{2}}{5 \times 10^{- 3}}$
$= 80.1 \times 10^{- 12} F = 80.1 p F$
$I = C \frac{d V}{d t} (∵ I = \frac{d q}{d t})$
$\Rightarrow \frac{d V}{d t} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{- 12}}$
$= 1.87 \times 10^{9} V s^{- 1}$
Displacement current is equal to conduction current $i . e ., 0.15 A$ .

PHXI15:WAVES

358751 In order to establish an instantaneous displacement current of $1 m A$ in the space between the parallel plates of $4 μ F$ capacitor what should be the rate of change of potential difference ?

1

350 V s^{- 1}

2

250 V s^{- 1}

3

100 V s^{- 1}

4

190 V s^{- 1}

Explanation:

$I_{D} = I_{C} = \frac{d}{d t} C V = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{4 \times 10^{- 6}} = 250 V s^{- 1}$

PHXI15:WAVES

358752 The area of each plate of a parallel plate condenser is $144 c m^{2}$ . The electrical field in the gap between the plates changes at the rate of $10^{12} V m^{- 1} s^{- 1}$ . The displacement current is

1

\frac{1}{10 π A}

2

\frac{0.4}{π} A

3

\frac{4}{π} A

4

\frac{40}{π} A

Explanation:

$I_{d} = ε_{0} A \frac{d E}{d t} = \frac{0.4}{π} A$

PHXI15:WAVES

358748 A parallel plate capacitor has a capacitance $C = 200 p F .$ It is connected to $230 V$ ac supply with an angular frequency $300 r a d / s .$ The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

1

13.8 μ A

and

138 μ A

2

14.3 μ A

and

143 μ A

3

13.8 μ A

and

13.8 μ A

4

1.38 μ A

and

1.38 μ A

Explanation:

$C = 200 p F, ω = 300 r a d / s, V = 230 V$
supporting img

$I = \frac{V}{X_{C}} = ω C V = 300 \times 200 \times 10^{- 12} \times 230$
$I = 13.8 μ A$
The magnitude of conduction current and displacement current is same.

JEE - 2024

PHXI15:WAVES

358749 An alternating voltage of amplitude $40 V$ and frequency $4 k H z$ is applied directly across the capacitor of $12 μ F .$ The maximum displacement current between the plates of the capacitor is nearly

1

13 A

2

8 A

3

10 A

4

12 A

Explanation:

Given, $V_{0} = 40 V,$ $f = 4 k H z = 4 \times 10^{3} H z$ $C = 12 \times 10^{- 6} F$
Voltage at any time ' $t$ ' is given by,
$V = V_{0} \sin (ω t) = V_{0} \sin (2 π f t)$
$∴ V = 40 \sin (2 π \times 4 \times 10^{3})$
$= 40 \sin (8 π \times 10^{3} t) V$
Displacement current, $I_{d} =$ Conduction current, $I_{C}$
$∴ I_{d} = C \frac{d V}{d t}$
$= 12 \times 10^{- 6} \times [40 \times 8 π \times 10^{3} \cos (8 π \times 10^{3} t)]$
$= 12.05 \cos (8 π \times 10^{3} t)$
$∴$ Maximum displacement current,
${(I_{d})}_{max} = 12.05 A (∴ For {(I_{d})}_{max,} \cos θ = 1)$
So, option (4) is correct

JEE - 2024

PHXI15:WAVES

358750 Figure shows a capacitor made of two circular plates each of radius $12 c m$ , separated by $5.0 m m$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$ , then which of the following is correct?
(Given: $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )
supporting img

1 The rate of change of potential difference between the plates is

1.87 \times 10^{9} {V s}^{- 1}

2 The displacement current across the plates is

0.30 A

.

3 The capacitance of the capacitor is

75 p F

.

4 None of these

Explanation:

Area, $A = π R^{2} = 3.14 \times (0.12)^{2} m^{2}$
$C = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 3.14 \times (0.12)^{2}}{5 \times 10^{- 3}}$
$= 80.1 \times 10^{- 12} F = 80.1 p F$
$I = C \frac{d V}{d t} (∵ I = \frac{d q}{d t})$
$\Rightarrow \frac{d V}{d t} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{- 12}}$
$= 1.87 \times 10^{9} V s^{- 1}$
Displacement current is equal to conduction current $i . e ., 0.15 A$ .

PHXI15:WAVES

358751 In order to establish an instantaneous displacement current of $1 m A$ in the space between the parallel plates of $4 μ F$ capacitor what should be the rate of change of potential difference ?

1

350 V s^{- 1}

2

250 V s^{- 1}

3

100 V s^{- 1}

4

190 V s^{- 1}

Explanation:

$I_{D} = I_{C} = \frac{d}{d t} C V = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{4 \times 10^{- 6}} = 250 V s^{- 1}$

PHXI15:WAVES

358752 The area of each plate of a parallel plate condenser is $144 c m^{2}$ . The electrical field in the gap between the plates changes at the rate of $10^{12} V m^{- 1} s^{- 1}$ . The displacement current is

1

\frac{1}{10 π A}

2

\frac{0.4}{π} A

3

\frac{4}{π} A

4

\frac{40}{π} A

Explanation:

$I_{d} = ε_{0} A \frac{d E}{d t} = \frac{0.4}{π} A$

PHXI15:WAVES

358748 A parallel plate capacitor has a capacitance $C = 200 p F .$ It is connected to $230 V$ ac supply with an angular frequency $300 r a d / s .$ The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

1

13.8 μ A

and

138 μ A

2

14.3 μ A

and

143 μ A

3

13.8 μ A

and

13.8 μ A

4

1.38 μ A

and

1.38 μ A

Explanation:

$C = 200 p F, ω = 300 r a d / s, V = 230 V$
supporting img

$I = \frac{V}{X_{C}} = ω C V = 300 \times 200 \times 10^{- 12} \times 230$
$I = 13.8 μ A$
The magnitude of conduction current and displacement current is same.

JEE - 2024

PHXI15:WAVES

358749 An alternating voltage of amplitude $40 V$ and frequency $4 k H z$ is applied directly across the capacitor of $12 μ F .$ The maximum displacement current between the plates of the capacitor is nearly

1

13 A

2

8 A

3

10 A

4

12 A

Explanation:

Given, $V_{0} = 40 V,$ $f = 4 k H z = 4 \times 10^{3} H z$ $C = 12 \times 10^{- 6} F$
Voltage at any time ' $t$ ' is given by,
$V = V_{0} \sin (ω t) = V_{0} \sin (2 π f t)$
$∴ V = 40 \sin (2 π \times 4 \times 10^{3})$
$= 40 \sin (8 π \times 10^{3} t) V$
Displacement current, $I_{d} =$ Conduction current, $I_{C}$
$∴ I_{d} = C \frac{d V}{d t}$
$= 12 \times 10^{- 6} \times [40 \times 8 π \times 10^{3} \cos (8 π \times 10^{3} t)]$
$= 12.05 \cos (8 π \times 10^{3} t)$
$∴$ Maximum displacement current,
${(I_{d})}_{max} = 12.05 A (∴ For {(I_{d})}_{max,} \cos θ = 1)$
So, option (4) is correct

JEE - 2024

PHXI15:WAVES

358750 Figure shows a capacitor made of two circular plates each of radius $12 c m$ , separated by $5.0 m m$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$ , then which of the following is correct?
(Given: $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )
supporting img

1 The rate of change of potential difference between the plates is

1.87 \times 10^{9} {V s}^{- 1}

2 The displacement current across the plates is

0.30 A

.

3 The capacitance of the capacitor is

75 p F

.

4 None of these

Explanation:

Area, $A = π R^{2} = 3.14 \times (0.12)^{2} m^{2}$
$C = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 3.14 \times (0.12)^{2}}{5 \times 10^{- 3}}$
$= 80.1 \times 10^{- 12} F = 80.1 p F$
$I = C \frac{d V}{d t} (∵ I = \frac{d q}{d t})$
$\Rightarrow \frac{d V}{d t} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{- 12}}$
$= 1.87 \times 10^{9} V s^{- 1}$
Displacement current is equal to conduction current $i . e ., 0.15 A$ .

PHXI15:WAVES

358751 In order to establish an instantaneous displacement current of $1 m A$ in the space between the parallel plates of $4 μ F$ capacitor what should be the rate of change of potential difference ?

1

350 V s^{- 1}

2

250 V s^{- 1}

3

100 V s^{- 1}

4

190 V s^{- 1}

Explanation:

$I_{D} = I_{C} = \frac{d}{d t} C V = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{4 \times 10^{- 6}} = 250 V s^{- 1}$

PHXI15:WAVES

358752 The area of each plate of a parallel plate condenser is $144 c m^{2}$ . The electrical field in the gap between the plates changes at the rate of $10^{12} V m^{- 1} s^{- 1}$ . The displacement current is

1

\frac{1}{10 π A}

2

\frac{0.4}{π} A

3

\frac{4}{π} A

4

\frac{40}{π} A

Explanation:

$I_{d} = ε_{0} A \frac{d E}{d t} = \frac{0.4}{π} A$

PHXI15:WAVES

358748 A parallel plate capacitor has a capacitance $C = 200 p F .$ It is connected to $230 V$ ac supply with an angular frequency $300 r a d / s .$ The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

1

13.8 μ A

and

138 μ A

2

14.3 μ A

and

143 μ A

3

13.8 μ A

and

13.8 μ A

4

1.38 μ A

and

1.38 μ A

Explanation:

$C = 200 p F, ω = 300 r a d / s, V = 230 V$
supporting img

$I = \frac{V}{X_{C}} = ω C V = 300 \times 200 \times 10^{- 12} \times 230$
$I = 13.8 μ A$
The magnitude of conduction current and displacement current is same.

JEE - 2024

PHXI15:WAVES

358749 An alternating voltage of amplitude $40 V$ and frequency $4 k H z$ is applied directly across the capacitor of $12 μ F .$ The maximum displacement current between the plates of the capacitor is nearly

1

13 A

2

8 A

3

10 A

4

12 A

Explanation:

Given, $V_{0} = 40 V,$ $f = 4 k H z = 4 \times 10^{3} H z$ $C = 12 \times 10^{- 6} F$
Voltage at any time ' $t$ ' is given by,
$V = V_{0} \sin (ω t) = V_{0} \sin (2 π f t)$
$∴ V = 40 \sin (2 π \times 4 \times 10^{3})$
$= 40 \sin (8 π \times 10^{3} t) V$
Displacement current, $I_{d} =$ Conduction current, $I_{C}$
$∴ I_{d} = C \frac{d V}{d t}$
$= 12 \times 10^{- 6} \times [40 \times 8 π \times 10^{3} \cos (8 π \times 10^{3} t)]$
$= 12.05 \cos (8 π \times 10^{3} t)$
$∴$ Maximum displacement current,
${(I_{d})}_{max} = 12.05 A (∴ For {(I_{d})}_{max,} \cos θ = 1)$
So, option (4) is correct

JEE - 2024

PHXI15:WAVES

358750 Figure shows a capacitor made of two circular plates each of radius $12 c m$ , separated by $5.0 m m$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$ , then which of the following is correct?
(Given: $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )
supporting img

1 The rate of change of potential difference between the plates is

1.87 \times 10^{9} {V s}^{- 1}

2 The displacement current across the plates is

0.30 A

.

3 The capacitance of the capacitor is

75 p F

.

4 None of these

Explanation:

Area, $A = π R^{2} = 3.14 \times (0.12)^{2} m^{2}$
$C = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 3.14 \times (0.12)^{2}}{5 \times 10^{- 3}}$
$= 80.1 \times 10^{- 12} F = 80.1 p F$
$I = C \frac{d V}{d t} (∵ I = \frac{d q}{d t})$
$\Rightarrow \frac{d V}{d t} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{- 12}}$
$= 1.87 \times 10^{9} V s^{- 1}$
Displacement current is equal to conduction current $i . e ., 0.15 A$ .

PHXI15:WAVES

358751 In order to establish an instantaneous displacement current of $1 m A$ in the space between the parallel plates of $4 μ F$ capacitor what should be the rate of change of potential difference ?

1

350 V s^{- 1}

2

250 V s^{- 1}

3

100 V s^{- 1}

4

190 V s^{- 1}

Explanation:

$I_{D} = I_{C} = \frac{d}{d t} C V = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{4 \times 10^{- 6}} = 250 V s^{- 1}$

PHXI15:WAVES

358752 The area of each plate of a parallel plate condenser is $144 c m^{2}$ . The electrical field in the gap between the plates changes at the rate of $10^{12} V m^{- 1} s^{- 1}$ . The displacement current is

1

\frac{1}{10 π A}

2

\frac{0.4}{π} A

3

\frac{4}{π} A

4

\frac{40}{π} A

Explanation:

$I_{d} = ε_{0} A \frac{d E}{d t} = \frac{0.4}{π} A$

PHXI15:WAVES

358748 A parallel plate capacitor has a capacitance $C = 200 p F .$ It is connected to $230 V$ ac supply with an angular frequency $300 r a d / s .$ The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

1

13.8 μ A

and

138 μ A

2

14.3 μ A

and

143 μ A

3

13.8 μ A

and

13.8 μ A

4

1.38 μ A

and

1.38 μ A

Explanation:

$C = 200 p F, ω = 300 r a d / s, V = 230 V$
supporting img

$I = \frac{V}{X_{C}} = ω C V = 300 \times 200 \times 10^{- 12} \times 230$
$I = 13.8 μ A$
The magnitude of conduction current and displacement current is same.

JEE - 2024

PHXI15:WAVES

358749 An alternating voltage of amplitude $40 V$ and frequency $4 k H z$ is applied directly across the capacitor of $12 μ F .$ The maximum displacement current between the plates of the capacitor is nearly

1

13 A

2

8 A

3

10 A

4

12 A

Explanation:

Given, $V_{0} = 40 V,$ $f = 4 k H z = 4 \times 10^{3} H z$ $C = 12 \times 10^{- 6} F$
Voltage at any time ' $t$ ' is given by,
$V = V_{0} \sin (ω t) = V_{0} \sin (2 π f t)$
$∴ V = 40 \sin (2 π \times 4 \times 10^{3})$
$= 40 \sin (8 π \times 10^{3} t) V$
Displacement current, $I_{d} =$ Conduction current, $I_{C}$
$∴ I_{d} = C \frac{d V}{d t}$
$= 12 \times 10^{- 6} \times [40 \times 8 π \times 10^{3} \cos (8 π \times 10^{3} t)]$
$= 12.05 \cos (8 π \times 10^{3} t)$
$∴$ Maximum displacement current,
${(I_{d})}_{max} = 12.05 A (∴ For {(I_{d})}_{max,} \cos θ = 1)$
So, option (4) is correct

JEE - 2024

PHXI15:WAVES

358750 Figure shows a capacitor made of two circular plates each of radius $12 c m$ , separated by $5.0 m m$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$ , then which of the following is correct?
(Given: $ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ )
supporting img

1 The rate of change of potential difference between the plates is

1.87 \times 10^{9} {V s}^{- 1}

2 The displacement current across the plates is

0.30 A

.

3 The capacitance of the capacitor is

75 p F

.

4 None of these

Explanation:

Area, $A = π R^{2} = 3.14 \times (0.12)^{2} m^{2}$
$C = \frac{ε_{0} A}{d} = \frac{8.85 \times 10^{- 12} \times 3.14 \times (0.12)^{2}}{5 \times 10^{- 3}}$
$= 80.1 \times 10^{- 12} F = 80.1 p F$
$I = C \frac{d V}{d t} (∵ I = \frac{d q}{d t})$
$\Rightarrow \frac{d V}{d t} = \frac{I}{C} = \frac{0.15}{80.1 \times 10^{- 12}}$
$= 1.87 \times 10^{9} V s^{- 1}$
Displacement current is equal to conduction current $i . e ., 0.15 A$ .

PHXI15:WAVES

358751 In order to establish an instantaneous displacement current of $1 m A$ in the space between the parallel plates of $4 μ F$ capacitor what should be the rate of change of potential difference ?

1

350 V s^{- 1}

2

250 V s^{- 1}

3

100 V s^{- 1}

4

190 V s^{- 1}

Explanation:

$I_{D} = I_{C} = \frac{d}{d t} C V = C \frac{d V}{d t}$
$∴ \frac{d V}{d t} = \frac{I_{D}}{C} = \frac{10^{- 3}}{4 \times 10^{- 6}} = 250 V s^{- 1}$

PHXI15:WAVES

358752 The area of each plate of a parallel plate condenser is $144 c m^{2}$ . The electrical field in the gap between the plates changes at the rate of $10^{12} V m^{- 1} s^{- 1}$ . The displacement current is

1

\frac{1}{10 π A}

2

\frac{0.4}{π} A

3

\frac{4}{π} A

4

\frac{40}{π} A

Explanation:

$I_{d} = ε_{0} A \frac{d E}{d t} = \frac{0.4}{π} A$

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