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Chapter 17: Problem 37

A parallel plate capacitor made to circular plates each of radius \(R=6 \mathrm{~cm}\) has capacitance \(C=100 \mathrm{pF}\). The capacitance is connected to a \(230 \mathrm{~V}\) AC supply with an angular frequency of \(300 \mathrm{rad} / \mathrm{s}\). The rms value of conduction current will be (A) \(5.7 \mu \mathrm{A}\) (B) \(6.3 \mu \mathrm{A}\) (C) \(9.6 \mu \mathrm{A}\) (D) \(6.9 \mu \mathrm{A}\)

Short Answer

Expert verified
The RMS value of the conduction current in the given parallel plate capacitor is (D) \(6.9 \mu A\).

Step by step solution

01

Calculate the capacitive reactance

: The formula for capacitive reactance (Xc) is given by: \[X_c = \frac{1}{\omega C}\] where 蠅 is the angular frequency and C is the capacitance. Plug in the given values: 蠅 = 300 rad/s, C = 100 pF = 100 脳 10鈦宦孤� F.
02

Find the capacitive reactance

: Using the formula from Step 1: \[X_c = \frac{1}{(300)(100 脳 10^{-12})}\] \[X_c = \frac{1}{3 脳 10^{-8}}\] \[X_c = \frac{1}{3} 脳 10^8 \Omega\]
03

Calculate the RMS current

: We can now use Ohm's Law to calculate the RMS current (I_rms): \[I_{rms} = \frac{V_{rms}}{X_c}\] where V_rms is the RMS voltage. Since the given voltage is 230 V, and the voltage supplied is an RMS value, we have V_rms = 230 V.
04

Find the RMS current

: Plug in values into the formula from Step 3: \[I_{rms} = \frac{230}{\frac{1}{3} \times 10^8}\] \[I_{rms} = 230 \times \frac{1}{33 \times 10^6}\] \[I_{rms} = 230 \times \frac{1}{33} \times 10^{-6} A\] \[I_{rms} = 6.97 \times 10^{-6} A\] Since, the answer is required in microamperes: \[I_{rms} = 6.97 \mu A\] We find that the RMS value of the conduction current is closest to the option (D) \(6.9 \mu A\). Therefore, the correct answer is (D) \(6.9 \mu A\).

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