Question

Some other capacitor problems, A. 5.0 mu F and a 15.0 mu F capacitor are connected in series and then this series combo is connected in parallel to a 25.0 mu F capacitor. What is the equivalent capacitance? What would be the voltage needed on a 10.0 F capacitor for it to store as much energy as a 1000 kg car traveling at 30.0 m/s? Five capacitors are connected in parallel, and then charged by an ideal battery of some voltage V, until each has charge q. If one of these three capacitors is then removed from the circuit, what would be the resulting charge on each remaining capacitor in the circuit?

Answer 1

A.

28.75 μF

B.

300 Volts

C.

q

Step-by-step explanation:

A.

`C_(1)` = 5 μF

`C_(2)` = 15 μF

`C_(s)` = Series combination of
`C_(1)` and
`C_(2)`

Series combination of
`C_(1)` and
`C_(2)` is given as

`C_(s)=(C_(1)C_(2))/(C_(1)+ C_(2))`

`C_(s)=((5)(15))/(5 + 15)`

`C_(s)` = 3.75 μF

`C_(3)` = 25 μF

`C_(eq)` = Equivalent capacitance

Equivalent capacitance is given as

`C_(eq)` =
`C_(s)` +
`C_(3)`

`C_(eq)` = 3.75 + 25

`C_(eq)` = 28.75 μF

B.

C = capacitance of the capacitor = 10 F

V = Voltage needed on the capacitor

m = mass of the car = 1000 kg

v = speed of the car = 30 m/s

Using conservation of energy

Energy stored by capacitor = Kinetic energy of car

(0.5) C V² = (0.5) m v²

(10) V² = (1000) (30)²

V = 300 Volts

C.

C = capacitance of each capacitor

q = charge stored by capacitor

V = battery voltage

If one of the capacitor is removed, the potential difference across each capacitor is same as battery voltage. hence the resulting charge on each capacitor remains the same as "q"