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1) Six 9.3??F capacitors are connected in parallel. What is the equivalent capac

ID: 2280426 • Letter: 1

Question

1) Six 9.3??F capacitors are connected in parallel.

What is the equivalent capacitance?

Ceq = ?F


What is their equivalent capacitance if connected in series?

Ceq = ?F


2) A circuit contains a single 260?pF capacitor hooked across a battery. It is desired to store three of two capacitors by adding a single capacitor to this one.


What would its value be?

C = pF


3) A power supply has a fixed output voltage of 12.0 V, but you need VT=5.0V for an experiment.

R2 =   ?


What will the terminal voltage VT be if you connect a load to the 5.0V terminal, assuming the load has a resistance of 4.0? ?

VT =   V

F capacitors are connected in parallel. What is the equivalent capacitance? What is their equivalent capacitance if connected in series? A circuit contains a single 260?pF capacitor hooked across a battery. It is desired to store three of two capacitors by adding a single capacitor to this one. What would its value be? A power supply has a fixed output voltage of 12.0 V, but you need VT=5.0V for an experiment. Using the voltage divider shown in the figure , what should R2 be if R1 is 16.0? What will the terminal voltage VT be if you connect a load to the 5.0V terminal, assuming the load has a resistance of 4.0? ?

Explanation / Answer

a) Parallel
======
If all the
capacitors are in parallel, the result is as if they were resistors in series. Since they are all the same, you get 6*9.3 uF = 55.8uF

Series
=====
If all the capacitors are in series, the result is like resistors in parallel. Since they are all the same, you get

9.3 uF/6 = 1.55 uF

b) in a given time, this battery can supply a fixed charge (Q)
energy stored at constant charge is
E = Q^2/2C
it reveals that for storing 3 times higher energy, final capacitance of new elements (
C1 & C2) [in the denominator] must decrease
>> the combination with can do this is > series combination of C1 & C2
1/C(final) = 1/C1 + 1/C2 = [C1+C2]/C1C2
----------------------------
given >
3 E (old) = E(new)
3 Q^2/2C1 = [Q^2/2C(final)]
3/C1 = [C1+C2]/C1C2
3C2 = [C1+C2]
2C2 = C1
C2 = C1/2 = 260/2 = 130 pF
>>>>>>>>
so connect 130 pF capacitor in series with given capacitor