Part 2 of 2

An amazing truth of electronics: When 2 capacitors are connected, either in series or parallel, it is possible to replace them with a single capacitance which will behave the same as the combination.

Try it with two capacitors in parallel:

We know the total voltage, V is the same across C₁ and C₂.

Therefore, q₁ = C₁V and q₂ = C₂V. The total charge delivered to the combined capacitance is q = q₁ + q₂. We can say, therefore, that

	q = C1V + C2V = (C1 + C2) V
but	q = CV where C is the combined capacitance.
Therefore	CV = (C1 + C2) V or C = C1 + C2

Now try it in series:

We know the total charge delivered to the plate of the first capacitor is the same as the charge on the second capacitor. Thus the total charge for the combined capacitance is identically equal to each individual charge on the string of series capacitors.

We may say, then, that V₁ = q/C₁ and V₂ = q/C₂

But the combined potential is V = V1 + V2 and V = q/C where C is the combined capacitance. Thus q/C = q/C₁ + q/C₂ and so:

1/C = 1/C₁ + 1/C₂

16. Find the combined capacitance of the circuits shown:

a) For parallel: C = C1 + C2 C = 12 + 7 C = 19 µF

b) For series: 1/C = 1/C1 + 1/C2 1/C = 1/9 + 1/4 C = 2.77 µF

c) 1/C = 1/3 + 1/7 + 1/5 C = 1.48 µF

d) First, combine the parallel caps: C = 7 + 9 + 2 C = 18 1/C = 1/14 + 1/18 C = 7.87 µF

17. What is the total charge on the combined capacitance? What is the charge on the 4 µF capacitor?

a) q = CV 1/C = 1/8 + 1/4 C = 2.67 µF q = (2.67 µ)(32) q = 85.3 µcoul

b) Also 85.3 µcoul of course

18. What is the potential on the 15 µF capacitor?

As in #17: 1/C = 1/6 + 1/15 + 1/9 C = 2.90 µF q = CV q = (2.90 µ)(16) q = 46.5 µcoul This must be q15 also, so: V15 = q/C V15 = (46.5 µ)/(15 µ) V15 = 3.10 V

19. What is the potential on the 11 µF capacitor?

1/C = 1/11 + 1/(8 + 15) C = 7.44 µF qtotal = CV qtotal = (7.44 µ)(20) qtotal = 149 µcoul But the total charge must reside on the 11µF, so: V = q/C V = (149 µ)/(11 µ) V = 13.5 V

20. What is the charge on the 6 µF capacitor?

q = CV q = (6 * 10-6)(9) q = 54.0 * 10-6 coul

21. What is the potential across the whole combined capacitance?

1/C = 1/9 + 1/16 + 1/100 + 1/22 C = 4.37 µF qtotal = q16 V = q/C V = (123 µcoul)/(4.37 µF) V = 28.2 V

22. What is the combined capacitance?

1/C1 = 1/22 + 1/15 + 1/30 C1 = 6.88 µF 1/C2 = 1/11 + 1/7 C2 = 4.28 µF 1/C = 1/37 + 1/(6.88 + 4.28) C = 8.57 µF

23. Again, what is the combined capacitance here?

1/C1 = 1/11 + 1/15 C1 = 6.35 µF C1 + 6 = 12.35 µF 1/CT = 1/12 + 1/12.35 + 1/20 C = 4.67 µF

24. What is the charge on the 8 µF capacitor?

V = q/C V = (10 * 10-6)/(16 * 10-6) V = .625 V across the 16 µF So across the 8 µF, V = .625 V q8 = CV q8 = (8 * 10-6)(.625) q8 = 5.00 µC

25. The charge on the 6 µF capacitor is 4 * 10^-5 C. What is the potential across the combined capacitance?

As in #24: V = q/C V = (4 * 10-5)/(6 * 10-6) V = 6.67 V across the 6µF So across the 11 µ, V = 6.67 q11 = CV q11 = (11 µF)(6.67) q11 = 73.3 µcoul Hence the total charge on the 11 and 6 combined is: Total charge = (4 * 10-5) + (73.3 * 10-6) Total charge = 11.3 * 10-5 coul This must equal q on the 8 µF, so: V8 = q/C V8 = (11.3 * 10-5)/(8 * 10-6) V8 = 14.1 V Vtotal = 6.67 + 14.1 Vtotal = 20.8 V

26. The potential across the 8 µF capacitor is 12 V. What is the charge on the 15 µF capacitor?

On the 8 µF: q = CV q = (8 µ)(12) q = 96 µC The combined charge on the 15 µF and 30 µF, then must be 96 µC. Think about it - 15 µF is 1/3 of the combined parallel capacitance and 30µF is 2/3 of the combined value. They must share 96 µC, so the 15 µF must get 1/3 or 32 µC on it.

By now you've done a lot of capacitor problems, some of which are fairly involved. Rather than conclude the chapter with super-challengers, I'd like to end by reinforcing those ideas that are basic to gaining a physical feel for what it's all about.

27. The charge on the 7 µF is 3 * 10^-4 C. What is the charge on the 14 µF?

3 * 10-4 C note: each capacitor pushes a charge onto the next, equal to the one it received.

28. The charge on C₁ is 7 * 10^-5 C, while C₂ carries 2 * 10^-5 C. What is the charge on C₃?

q2 + q3 = q1 (2 * 10-5) + q3 = 7 * 10-5 q3 = 5 * 10-5 C

29. The potential on C₁ is 80 V, on C₂ is 20 V. What are the potentials across C₃ and C₄?

V1 = V3 V3 = 80 V V2 + V4 = V1 V4 = V1 - V2 V4 = 80 - 20 V4 = 60 V

30. Find the combined capacitance of each network. All capacitances are in µF.

a) 1/C = 1/12 + 1/(3 + 5) C = 4.8 µf

b) 1/C1 = 1/18 + 1/12 C1 = 7.2 1/C = 1/20 + 1/(6 + 7.2) C = 7.95 µf

c) 1/C1 = 1/16 + 1/(5 + 4) C1 = 5.76 C = 5.76 + 6 C = 11.76 µf

d) 1/C1 = 1/18 + 1/10 + 1/12 C1 = 4.29 1/C2 = 1/6 + 1/10 C2 = 3.75 1/C = 1/15 + 1/(4.29 + 3.75) C = 5.24 µf

31. In each circuit find the unknown charge.
a)
b)
c)
d)
e)
f)

a) q = 8 µC	b) q = CV q = (8µF)(32 V) q = 256 µC	c) q = C1 note: V is applied directly across C1
d) q = V(1)/[(1/C1) + (1/C2 + (1/C3) + (1/C4)]	e) V = q/C V = (9µC)/(15µF) q = CV q = (27µ)(9/15) q = 16.2 µC	f) q1 = CV q1 = (8 µ)(11/9) q1 = 22.0 µC q2 = 22 µ + 11 µ q2 = 33.0 µC q3 = q2 q3 = 33.0 µC

32. Find the potentials:
a)
b)
c)
d)
e)

a) V = 20 V	b) V7 = q/C V7 = (6 µ)/(7 µ) V7 = .857 V V8 = q/C V8 = (6 µ)/(8 µ) V8 = .750 V	c) q12/(5) = (12 µ)/(8 µ) q12 = 7.5 µC V7 = q7/C7 V7 = (q8 + q12)/C7 V7 = (5 µ + 7.5 µ)/(7 µ) V7 = 1.79 V
d) V15 = V25 V15 = q/C V15 = (11 µ)/(25 µ) V15 = .440 V	e) V = q/CT V = (6 µ)/(10.98 µF) V = .546 V

33. Find the equivalent capacitances:
a)
b)
c)

a) CT = C1 + C2 + C3	b) CT = (1)/[(1/C1) + (1/C2)]
c) CT = (1)/[(1/C1) + (1)/(C2 + C3)]

34. Given that q₉ = 24 µC, find V₉, q₁₅, q₁₈, V_AB, q₁₀, q₅₀, and V_total.

V9 = q/C V9 = (24 µ)/(9 µ) V9 = 2.67 V q15 = CV q15 = (15 µ)(2.67) q15 = 40.0 µC q18 = 24 µ + 40 µ q18 = 64.0 µC VAB = q/C VAB = (64 µ)/(1/[(1/18 µ) + [1/(15 µ + 9 µ)]] VAB6.22 V q10 = CV q10 = [(1)/[(1/10) + (1/20) + (1/12)]](6.22) q10 = 26.7 µC q50 = q10 + q18 q50 = 26.7 µ + 64 µ q50 = 90.7 µC VT = V50 + VAB VT = (q/C) + VAB VT = (90.7 µ)/(50 µ) + 6.22 VT = 8.03 V