Physics, 26.02.2020 20:00 coolkason

Two capacitors with capacitances of 1.0 m F and 0.50 m F, respectively, are connected in series. The system is connected to a 100-V battery. What electrical potential energy is stored in the 1.0- m F capacitor?

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Answer from: stheo

The energy stored in the 1.0mF capacitor is 9.0*10^-5J

Data given;

= 1.0mFc

= 0.50mFV = 100VCapacitor

Since the capacitor are in series, the total capacitance is;

$C = \frac{1}{c_1} + \frac{1}{c_2}$

substitute in the values and solve.

$C = \frac{1}{1} + \frac{1}{0.5}\\C = 3mF$

The capacitance in the capacitor is 3mF.

Potential Energy

This is the energy stored in the capacitor.

$U= \frac{1}{2} q^2C$

U = potential energy

q = charge

C = capacitor.

But the charge of a capacitor is given as

$q = cV\\q = (3*10^-^3)*100\\q = 0.3C$

The charge on the capacitor is 0.3C

We can proceed to use this formula and solve for the potential energy stored in the 1.0mF capacitor.

$U = \frac{1}{2} q^2C\\U = \frac{1}{2}* 0.3^2*1.0^-^3\\U = 9.0*10^-^5J$

From the calculations above, the potential energy stored in the 1mF capacitor is 9.0*10^-^5J .

Learn more on capacitor here;

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Answer from: BrianKeokot4534

Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J

Explanation:

Series capacitor: The ending point of a capacitor is the starting point of other capacitor.

If C₁ and C₂ are connected in series then the equivalent capacitance is C.

where $\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}$

Given that,

C₁ = 1.0 mF=1.0×10⁻³F and C₂ = 0.50mF=0.50×10⁻³F

If C is equivalent capacitance.

Then $\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}$

$\Rightarrow \frac{1}{C} =\frac{3}{1}$

$\Rightarrow C=\frac{1}{3}$ mF

Again given that the system is connected to a 100-v battery.

We know that

q=Cv

q= charge

C= capacitor

v= potential difference

Therefore

$q=(\frac{1}{3} \times 10^{-3}\times10) C$

$=\frac{10^{-2}}{3} C$

The electrical potential energy stored in a capacitor can be expressed

$U=\frac{q^2C}{2}$

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

$U=\frac{q^2C_1}{2}$

$\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}$

$\Rightarrow U= 5.56\times 10^{-9}$ J

Answer from: Quest

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.

Answer from: Quest

Aliquid is an answer i hope this

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