Question

In the circuit below, VBAT= 17 V?

(Questions and picture below.)

Answer 1

a) What is the equivalent capacitance for the combination of the 6 and 3 μF capacitors?
Express your answer using three significant figures. Answer in μF.

b) What is the equivalent capacitance for the entire circuit? Answer in μF.

c) How much charge does the battery supply? Answer in μC.

d) What is the charge on the 6 μF capacitor? Answer in μC.

e) What is the voltage across the 6 μF capacitor? Answer in V.

f) What is the voltage across the 3 μF capacitor? Answer in V.

g) What is the charge on the 2μF capacitor? Answer in μC.

h) What is the total energy stored by the capacitors? Answer in J.

Answer 2

The capacitance in the path with them in series is
1/C = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 so C = 2 uF

The capacitance for the combination of that path and the one it is in parallel with is C' = C + 2 uF = 4 uF.

Q = C' V for the combined total circuit.

The 2 uF capacitor has V across it so Q' = (2 uF) V.

Energy = (1/2) C' V^2

Answer 3

Alright so:
a) 2 uF
b) 4 uF
Ah, I didn't realize that capacitors had flipped equations versus capacitors. I just learned this in class, so that makes more sense.

Let me try to work out the others with the info you provided me.

Answer 4

c) Q = C' V
(4)(17) = 68 uF
Alright, got that right

Answer 5

d) Q = C' V
(6)(17) = 107 uF

Ahhh, that's too big...

Hm. Well, I guess that doesn't make sense anyway because C' is 2 and I can't replace it.

But I was under the impression that the other equations in the list are for g) and h)?

Answer 6

I did not answer all the questions in list: 5 answers, 8 questions. I think the principles needed are there, tho.

Answer 7

Alright, I managed to find the majority:

d) 34μC
e) V = Q/C = 34e-6/6e-6 = 5.67V
f) V = Q/C = 34e-6/3e-6 = 11.333V
g) 34μC

Answer 8

h) E = 1/2*4e-6*17² = 578μJ > 5.78^-4 J

Okay, got it!