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
b. Enegy storedinthecapacitor U = (1/2)*C * V2
= 0.5*79 * 10-6 * 172
= 1.14*10-2 J
c. Maximumenergyinindcutor = U
(1/2)*L*Imax2 = 1.14*10-2 0.5 * 66*10-3*Imax2 = 1.14*10-2
Imax = â(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) ???