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schepotkina [342]
3 years ago
5

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 fr

equency 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.
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

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

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