Question 412: AC fundamentals

Question

A parallel plate capacitor having plates of radius 6 cm has capacitance 100 pF. It is connected to 230 V a.c. supply with angular frequency 300 rad/s. The r.m.s. value of current is

Accepted Answer

To find the r.m.s. value of the current in a circuit containing a capacitor connected to an AC supply, we use the relation $$I_{rms} = V_{rms} 	imes \omega C$$ where $$I_{rms}$$ is the root mean square (r.m.s.) current, $$V_{rms}$$ is the root mean square voltage (which is equivalent to the given voltage for an AC circuit, in this case, 230 V), $$\omega$$ is the angular frequency, and $$C$$ is the capacitance of the capacitor.
Given:
Capacitance, $$C = 100 \, pF = 100 	imes 10^{-12} \, F$$
Angular frequency, $$\omega = 300 \, rad/s$$
AC supply voltage, $$V = 230 \, V$$
Therefore, we can substitute the given values into our formula:
$$I_{rms} = V_{rms} 	imes \omega C$$
$$I_{rms} = 230 	imes 300 	imes 100 	imes 10^{-12}$$
$$I_{rms} = 6.9 	imes 10^{-6} \, A$$
Thus, the r.m.s. current is $$6.9 	imes 10^{-6}$$ A, which matches Option A.

Answer

$$6.9	imes10^{-6}$$ A

Answer

$$2.3	imes10^{-5}$$ A

Answer

$$6.9	imes10^{-5}$$ A

Answer

$$6.9	imes10^{-7}$$ A

AC fundamentals

A parallel plate capacitor having plates of radius 6 cm has capacitance 100 pF. It is connected to 230 V a.c. supply with angular frequency 300 rad/s. The r.m.s. value of current is

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