Notice that the equivalent capacitance is less than that of any of the individua

Question

Notice that the equivalent capacitance is less than that of any of the individual capacitors. The relationship C = Q/?V can be used to find the voltage drops on the other capacitors, just as inpart (c).

QUESTION Over which capacitor is the voltage drop the smallest?

the 3.0-µF capacitor

the 6.0-µF capacitor    

the 12-µF capacitor

the 24-µF capacitor

Over which capacitor is the voltage drop the largest?

the 3.0-µF capacitor

the 6.0-µF capacitor    

the 12-µF capacitor

the 24-µF capacitor

PRACTICE IT

Four capacitors are connected in series with a battery, as in the figure below, where C1 = 3.60 µF, C2 = 6.80 µF, C3 = 12.4 µF, C4 = 24.9 µF,V = 18.2 V.

(a) Calculate the capacitance of the equivalent capacitor.
µF

(b) Compute the charge on C3.
µC

(c) Find the voltage drop across C3.
V

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise. C4 is removed from the circuit, leaving only three capacitors in series.

(a) Find the equivalent capacitance.
Ceq =  µF

(b) Find the charge on C2.
Q =  µC

(c) Find the voltage drop across C2.
?V =  V

Explanation / Answer

Answer:

We need a diagram and previous information to answer the first two questions. Here are the solutions for Next questions.

PRACTICE IT:

(a) In series combination,1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4 , therefore,

Ceq = C1C2C3C4/(C2C3C4 + C1C3C4+C1C2C4+C1C2C3)

= (3.60 µF x 6.80 µF x 12.4 µF x 24.9 µF) / (6.80 µF x 12.4 µF x 24.9 µF) + (3.60 µFx 12.4 µF x 24.9 µF) + (3.60 µF x 6.80 µF x24.9 µF ) + (3.60 µF x 6.80 µF x 12.4 µF)

= (7.55 x 10-21 F4) / (2.09 x 10-15 F3) + (1.11 x 10-15 F3) + (0.609 x 10-15 F3 ) + (0.303 x 10-15 F3)

Therefore, Ceq = 1.83 x 10-6 F

(b) Total charge in the circuit Q = Ceq V = (1.83 x 10-6 F) (18.2 V) = 33.30 x 10-6 C.

Charge is same for all capacitors in series combination, therefore Q3 = Q = 33.30 x 10-6 C

(c) Voltage drop across the capacitor C3 is V3 = Q3/C3 = (33.30 x 10-6 C) / (12.4 x 10-6 F) = 2.68 V.

Exercise:

C4 is removed from the circuit, leaving only three capacitors in series. then

(a) Ceq = C1C2C3/(C2C3 + C1C3+C1C2)

= (3.60 µF x 6.80 µF x 12.4 µF ) / (6.80 µF x 12.4 µF ) + (3.60 µFx 12.4 µF ) + (3.60 µF x 6.80 µF )

= (0.303 x 10-15 F3) / (0.0843 x 10-9 F2) + (0.044 x 10-9 F2) + (0.024 x 10-9 F2) = 2.00 X 10-6 F.

(b) Total charge Q = Ceq V = (2.00 X 10-6 F) (18.2 V) = 36.4 x 10-6 C

Therefore, charge on C2 is Q2 = Q = 36.4 x 10-6 C.

(c) Voltage drop across C2 is V2 = Q2/C2 = (36.4 x 10-6 C) / (6.80 x 10-6 F) = 5.35 V.

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