Question

A 63.0 \mu F capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 4.00 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 15.0 A. As the generator frequency is increased, at what frequency will the fuse burn out

Answer 1

`f=9.5\ KHz`

Step-by-step explanation:

AC Circuit

When connected to an AC circuit, the capacitor acts as an impedance of module

`\displaystyle Z=(1)/(wC)`

Where w is the angular frequency of the power source and C is the capacitance.

If the capacitor is the only element connected to a circuit, then the Ohm's law establishes that

`V=Z.I`

Where V and I are the rms voltage and current respectively. Replacing the value of Z, we have

`\displaystyle V=(I)/(wC)`

Solving for w

`\displaystyle w=(I)/(VC)`

The question provides us the following values

`C=63\ \mu F=63\cdot 10^(-6)\ F`

`V=4\ Volt`

`I=15\ A`

Plugging in the values

`\displaystyle w=(15)/(4\cdot 63\cdot 10^(-6))`

`w=59523.81\ rad/s`

Since

`w=2\pi f`

Then

`\displaystyle f=(59523.81)/(2\pi)=9473.51\ Hz`

`f=9.5\ KHz`