Question

9. The peak current through a capacitor is 12.0 mA .

Part A)

What is the current if the emf frequency is doubled? Express answer in Milliamperes.

Part B)

What is the current if the emf peak voltage is doubled (at the original frequency)? Express answer in milliamperes.

Part C)

What is the current if the frequency is halved and, at the same time, the emf is doubled? Express answer in milliamperes

10. A 19

mH inductor is connected across an AC generator that produces a peak voltage of 9.00 V .

Part A)

What is the peak current through the inductor if the emf frequency is 100 Hz? Express answer in amperes.

Part B)

What is the peak current through the inductor if the emf frequency is 100 kHz? Express answer in amperes.

Answer 1

I = V/Xc

Xc =1/2*pi*f*C

so I =V/ (1/ 2*pi*f*C) = V*2*pi*f*C

so if f= > 2f then

A ) I' =V(2π(2f)C) = 2*I

= 2*12 = 24 mA

B)   V=>2V

then I′=(2V)(2πfC)=2V(2πfC)​ = 2*I = 2*12 = 24 mA

C) V= >2V anf f= >f/2

I′=V′(2πf′C)

= 2*V *2*π* f/2 *C = V*2*pi*f*C​ = I = 12 mA

10.​

a) Xl = 2*pi*f*L = 2*pi*100* 19*10^-3 = 11.9380521 ohm

I = V/Xl =9/11.9380521 = 0.753891835 A

b) Xl = 2*pi*f*L = 2*pi*100*10^3* 19*10^-3 =11938.0521 ohm

so I= 9/11938.0521 = 0.000753891835A = 0.75389 mA