A parallel plate condenser of capacity $100\,pF$ is connected to $230\,V$ of AC supply of $300\,rad{\operatorname{s} ^{ - 1}}$ frequency. The rms value of displacement current.
(A) $6.9\,\mu A$
(B) $2.3\,\mu A$
(C) $9.2\,\mu A$
(D) $4.6\,\mu A$
Answer
297k+ views
Hint The rms value of the current is determined by using the rms formula of the current and the rms value of the current is equal to the rms of the voltage divided by the inductive capacitance of the capacitor. Then the inductive capacitance is written as the product of the frequency and the capacitance.
Useful formula
The rms value of the current is given as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}$
Where, ${I_{rms}}$ is the rms value of the current, ${E_{rms}}$ is the rms value of the voltage and ${X_C}$ is the inductive capacitance.
Complete step by step answer
Given that,
The capacitance of the parallel plate is given as, $C = 100\,pF = 100 \times {10^{ - 12}}\,F$,
The rms value of the voltage is given as, ${E_{rms}} = 230\,V$,
The angular frequency is given as, $\omega = 300\,rad{\operatorname{s} ^{ - 1}}$.
Now,
The rms value of the current is given as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}\,....................\left( 1 \right)$
Now, the inductive capacitance is written as ${X_C} = \dfrac{1}{{\omega C}}$. By substituting this equation in the above equation, then the above equation is written as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
${I_{rms}} = {E_{rms}} \times \omega C$
By substituting the rms value of the voltage, angular frequency and the capacitance of the parallel plate in the above equation, then the above equation is written as,
${I_{rms}} = 230 \times 300 \times 100 \times {10^{ - 12}}$
By multiplying the terms in the above equation, then the above equation is written as,
${I_{rms}} = 6.9 \times {10^{ - 6}}\,A$
Then the above equation is also be written as,
${I_{rms}} = 6.9\,\mu A$
Thus, the above equation shows the rms value of displacement current.
Hence, the option (A) is the correct answer.
Note The rms value of displacement current is directly proportional to the rms value of the voltage and inversely proportional to the inductive capacitance. As the rms value of the voltage increases, then the rms value of displacement current also increases.
Useful formula
The rms value of the current is given as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}$
Where, ${I_{rms}}$ is the rms value of the current, ${E_{rms}}$ is the rms value of the voltage and ${X_C}$ is the inductive capacitance.
Complete step by step answer
Given that,
The capacitance of the parallel plate is given as, $C = 100\,pF = 100 \times {10^{ - 12}}\,F$,
The rms value of the voltage is given as, ${E_{rms}} = 230\,V$,
The angular frequency is given as, $\omega = 300\,rad{\operatorname{s} ^{ - 1}}$.
Now,
The rms value of the current is given as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}\,....................\left( 1 \right)$
Now, the inductive capacitance is written as ${X_C} = \dfrac{1}{{\omega C}}$. By substituting this equation in the above equation, then the above equation is written as,
${I_{rms}} = \dfrac{{{E_{rms}}}}{{\left( {\dfrac{1}{{\omega C}}} \right)}}$
By rearranging the terms in the above equation, then the above equation is written as,
${I_{rms}} = {E_{rms}} \times \omega C$
By substituting the rms value of the voltage, angular frequency and the capacitance of the parallel plate in the above equation, then the above equation is written as,
${I_{rms}} = 230 \times 300 \times 100 \times {10^{ - 12}}$
By multiplying the terms in the above equation, then the above equation is written as,
${I_{rms}} = 6.9 \times {10^{ - 6}}\,A$
Then the above equation is also be written as,
${I_{rms}} = 6.9\,\mu A$
Thus, the above equation shows the rms value of displacement current.
Hence, the option (A) is the correct answer.
Note The rms value of displacement current is directly proportional to the rms value of the voltage and inversely proportional to the inductive capacitance. As the rms value of the voltage increases, then the rms value of displacement current also increases.
Recently Updated Pages
How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Hydrogen and Its Type Important Concepts and Tips for JEE Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

