Question 141: AC fundamentals

Question

An alternating voltage of amplitude $$40 \mathrm{~V}$$ and frequency $$4 \mathrm{~kHz}$$ is applied directly across the capacitor of $$12 \mu \mathrm{F}$$. The maximum displacement current between the plates of the capacitor is nearly :

Accepted Answer

Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.
The formula for the capacitive reactance $$X_C$$ of a capacitor is given by:
$X_C = \frac{1}{2\pi fC}$
where
$$f$$ is the frequency of the alternating voltage, and
$$C$$ is the capacitance of the capacitor.
Given:
The amplitude of the voltage ($$V_{max}$$) = $$40$$ V
Frequency ($$f$$) = $$4000$$ Hz (or $$4 \mathrm{~kHz}$$)
Capacitance ($$C$$) = $$12 \mu \mathrm{F} = 12 	imes 10^{-6} \mathrm{~F}$$
First, we calculate $$X_C$$:
$X_C = \frac{1}{2\pi(4000)(12 	imes 10^{-6})} \approx \frac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \frac{1}{96\pi \cdot 10^{-3}}$
$X_C \approx \frac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \frac{1}{0.30144} \approx 3.316 \Omega$
Now, the maximum current ($$I_{max}$$) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance:
$I_{max} = \frac{V_{max}}{X_C}$
$I_{max} = \frac{40}{3.316} \approx 12.06 \mathrm{~A}$
Hence, the maximum displacement current between the plates of the capacitor is nearly:
Option D: 12 A

Answer

10 A

Answer

8 A

Answer

13 A

Answer

12 A

AC fundamentals

An alternating voltage of amplitude $40 \mathrm{~V} and frequency 4 \mathrm{~kHz} is applied directly across the capacitor of 12 \mu \mathrm{F}$. The maximum displacement current between the plates of the capacitor is nearly :

Choose the correct answer: