Question

In the diagram below the capacitor has a capacitance of79μF and the inductor has an inductance of 66 mH.Initiallythe capacitor has a voltage of 17 V. This circuit hasnoresistance.


A. How much charge does the capacitorinitiallyhave?
1 C

B. How much energy does the capacitorinitiallyhave?
2 J

C. What is the maximum current that willpassthrough the inductor?
3 A

Response Details:

a. Charge Q = C*V

= 79*10^-6 * 17

= 1.343*10^-3 C

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b. Enegy storedinthecapacitor U = (1/2)*C * V²

= 0.5*79 * 10^-6 * 17²

= 1.14*10^-2 J

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c. Maximumenergyinindcutor = U

(1/2)*L*I_max² = 1.14*10^-2 0.5 * 66*10^-3*I_max² = 1.14*10^-2

I_max = √(1.14*10^-2 / 3.30 * 10^-2)

= 0.5877 A

An object is held 99.5 cm away from a mirror. The mirror hasafocal length of 72.9 cm.

a.) Determine the image distance.
1 cm

b.) Determine the magnification.
2

c.) If the object is 9.8 cm long, determine the image height.
3 cm

d.)Which of the following are true about the image? Choose allthatapply.

4

The image is virtual. The image isupright. The image isinverted. The image isreal.

e.)Which of the following are true about the mirror? Choose allthatapply.

5

The mirror is concave. The mirror isconvex. The mirror isconverging. The mirror isdiverging.

Answer:

a) The image distance

q = pf / p -f =99.5*72.9 / 99.6-72.9 = 272.69cm

(b) Magnification m = -q/p = -272.69 / 99.5 =-2.74

(c) magnification m = hf /hi

-2.74 = hf /9.8

hf = -26.85cm

(d) for lens image is realandinverted

(e) ???