Physics, 02.02.2020 02:45 ashhleyjohnson

(a) you short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. if the current in the short circuit is measured to be 18 amperes, what is the internal resistance of the battery? (b) what is the power generated by the battery? w (c) how much energy is dissipated in the internal resistance every second? (remember that one watt is one joule per second.) j (d) this same battery is now connected to a 11 resistor. how much current flows through this resistor? a (e) how much power is dissipated in the 11 resistor? w (f) the leads to a voltmeter are placed at the two ends of the battery of this circuit containing the 11 resistor. what does the meter read?

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Answer from: yradf57441

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

P=VI

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

P=(20 V)(18 A)=360 W

The energy dissipated by the internal resistance is given by

E=Pt

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

E=(360 W)(1 s)=360 J

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

P=I^2 R

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

Answer from: Quest

gas and liquid

Answer from: Quest

answer;

d. are made mostly of liquid

explanation;

-the general composition and structure planets includes; they have a thick gas and liquid layer covering a small,solid core. mostly the gases are hydrogen and helium atmosphere and layers of liquid hydrogen. for instance neptune and uranus are examples of the outer planets, they have a core covered by ice and ammonia.

-neptune's atmosphere is hydrogen and helium with a bit of methane so there is no oxygen to breath and the interior is made up of frozen water and ammonia with a rock and iron core. thus, there is no liquid water or solid surface to survive.

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