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Question: A parallel plate condenser of capacity \(100\,pF\) is connected to \(230\,V\) of AC supply of \(300\...

A parallel plate condenser of capacity 100 pF100\,pF is connected to 230 V230\,V of AC supply of 300 rads⁡−1300\,rad{\operatorname{s} ^{ - 1}} frequency. The rms value of displacement current.
(A) 6.9 μA6.9\,\mu A
(B) 2.3 μA2.3\,\mu A
(C) 9.2 μA9.2\,\mu A
(D) 4.6 μA4.6\,\mu A

Explanation

Solution

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,
Irms=ErmsXC{I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}
Where, Irms{I_{rms}} is the rms value of the current, Erms{E_{rms}} is the rms value of the voltage and XC{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=100×10−12 FC = 100\,pF = 100 \times {10^{ - 12}}\,F,
The rms value of the voltage is given as, Erms=230 V{E_{rms}} = 230\,V,
The angular frequency is given as, ω=300 rads⁡−1\omega = 300\,rad{\operatorname{s} ^{ - 1}}.
Now,
The rms value of the current is given as,
Irms=ErmsXC ....................(1){I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_C}}}\,....................\left( 1 \right)
Now, the inductive capacitance is written as XC=1ωC{X_C} = \dfrac{1}{{\omega C}}. By substituting this equation in the above equation, then the above equation is written as,
Irms=Erms(1ωC){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,
Irms=Erms×ωC{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,
Irms=230×300×100×10−12{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,
Irms=6.9×10−6 A{I_{rms}} = 6.9 \times {10^{ - 6}}\,A
Then the above equation is also be written as,
Irms=6.9 μA{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.