Question

a. A 91.0-μF capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 2.40 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 11.0 A. As the generator frequency is increased, at what frequency will the fuse burn out?

b. Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.00 s. How much time does the driver of the car measure for his trip between the poles?

Answer 1

a) Lets,

Fuse will burn out when capacitor reactance Xc = V / I ......................(1)

where  The r.m.s voltage of the generator is 2.40 V

and  the r.m.s current is 11.0 A

put the value on equation number (1)

  

   Ohms

we know that,

  

The   Frequency will the fuse burn out is 8016 Hz

b) the speed of light in a vacuum has the hypothetical value of 18.0 m/s =C

A car is moving at a constant speed of 14.0 m/s = V

T = t/sqrt(1 – (V/C)^2) = 8.0 sec is the observer’s time interval,

we will calculate the value of solve for t = ?

Then, the driver's comparable interval when V/C,

  

From above equation,  T = t/sqrt(1 – (V/C)^2) = 8.0

  

  

  

  

Time does the driver of the car measure for his trip between the poles is 5.03 sec