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ICE Princess25 [194]
3 years ago
9

To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 1.6FF capac

itor charged to 24 VV, double the normal 12 VV provided by the car's battery. How much energy does the capacitor store at 12 V? How much energy does the capacitor store at 24 V?
Physics
2 answers:
AleksandrR [38]3 years ago
7 0

Answer:

Explanation:

Capacitance, C = 1.6 F

(a) V = 12 V

Energy stored in a capacitor,

U = 0.5 C V²

U = 0.5 x 1.6 x 12 x 12 = 115.2 J

(b) V = 24 V

Energy stored in a capacitor,

U = 0.5 C V²

U = 0.5 x 1.6 x 24 x 24 = 460.8 J

Hoochie [10]3 years ago
6 0

Answer:

Energy stored at 12 volt will be 115.2 J and energy stored at 24 volt will be 460.8 volt

Explanation:

We have given capacitance of capacitor C = 1.6 F

Normal voltage V = 12 volt

And system uses 2 volt to charge the system

We know that energy stored in the capacitor is equal to E=\frac{1}{2}CV^2

So energy stored at 12 volt will be equal to E=\frac{1}{2}\times 1.6\times 12^2=115.2j

And energy stored at 24 volt will be equal to E=\frac{1}{2}\times 1.6\times 24^2=460.8j

So energy stored at 12 volt will be 115.2 j and energy stored at 24 volt will be 460.8 j

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1 year ago
Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

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V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

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Answer:

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Answer:

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