In the diagram below the capacitor has a capacitance of 79 F and the inductor ha
ID: 1678146 • Letter: I
Question
In the diagram below the capacitor has a capacitance of 79F and the inductor has an inductance of 66 mH.Initially the capacitor has a voltage of 17 V. This circuit has noresistance.
A. How much charge does the capacitor initiallyhave?
1 C
B. How much energy does the capacitor initiallyhave?
2 J
C. What is the maximum current that will passthrough the inductor?
3 A
Response Details:
a. Charge Q = C* V
= 79* 10-6 * 17
= 1.343* 10-3 C
b. Enegy storedin thecapacitor U = (1/2)* C * V2
= 0.5* 79 * 10-6 * 172
= 1.14* 10-2 J
c. Maximum energyin indcutor = U
(1/2)* L *Imax2 = 1.14* 10-2 0.5 * 66* 10-3 *Imax2 = 1.14*10-2
Imax = (1.14* 10-2 / 3.30 * 10-2)
= 0.5877 A
An object is held 99.5 cm away from a mirror. The mirror has afocal 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 all thatapply.
4
The image is virtual. The image isupright. The image is inverted. The image isreal.
e.)Which of the following are true about the mirror? Choose all thatapply.
5
The mirror is concave. The mirror isconvex. The mirror is converging. The mirror isdiverging.
Answer:
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 = -26.85 cm
(d) for lens image is real andinverted
Explanation / Answer
Your answers for the inductor question is correct .. But in mirror problem , the object distance is taken to be negativeby convention . So put p=-99.5 cm to get the answers . Also as far as the mirror is concerned , here focal length ispositive , So from convention of mirrors , the mirror is convex andhence diverging .