Question

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to 32.6 kHz. 32.6 MHz. 5.20 kHz. 5.20 MHz. 32.6 Hz.

Answer 1

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance $C=8.5nF=8.5\times 10^{-9}f$

Potential difference across capacitor V = 12 volt

Current through capacitor $i=3.33mA=3.33\times 10^{-3}A$

Capacitive reactance will be equal to $X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm$

Capacitive reactance is equal to $X_c=\frac{1}{\omega C}$

$3603.60=\frac{1}{\omega\times 8.5\times 10^{-9}}$

$\omega =32647.091rad/sec$

$2\pi f=32647.091$

$f=5198.98Hz$

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer