Que es un patrimonio histórico Nacional

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

c)

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

3 years ago

A 190 g rock at 20°C is completely immersed in 600 g of water at 80°C. Which one of the following statements is true right after

Umnica [9.8K]

Rock is completely immersed in hot water. By the second law of thermodynamics, thermal energy or heat is transferred from substance with higher temperature to substance with lower temperature until they come to thermal equilibrium i.e. both at same temperature.

It is given here that rock is at 20°C which is at lower temperature than water at 80°C. ∴Heat or thermal energy flows from water to rock. So, right choice is-

A. The water gives the rock thermal energy and gets no thermal energy in return.

8 0

3 years ago

The four terrestrial planets are so called because they ?

velikii [3]

Well, you need no look further than the word "terrestrial" If you notice the beginning of the word you notice that it consists mostly of the word "terra" Terra by definition is just land. Due to the solid land of these 4 planets, they're called terrestrial planets, the other 4 aren't made of land but of gas which is why they aren't classified as terrestrial planets.

8 0

3 years ago

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A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

$\phi = 0.85$ rad = 48.72°

(a). Impedance of the circuit is given by

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

$Z = \frac{150}{2.75}$

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

$R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }$

Put the values of Z & $\phi$ in above formula we get

$R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }$

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0

3 years ago

Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a

irina1246 [14]

Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

I₁ = I₀ / 2

I₁ = 22.4 / 2

I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

I₂ = I₁ cos² θ

I₂ = 11.2 cos² 69

I₂ = 1.44 Lux

8 0

3 years ago

Que es un patrimonio histórico Nacional​

Que es un patrimonio histórico Nacional