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Lyrx [107]
3 years ago
6

What is approximate resistance between P and Q? a) 0.5 b)0.8 c)2.0 d)2.2 e)3.6

Physics
1 answer:
Rufina [12.5K]3 years ago
8 0

Answer:

0.5 Ohms

Explanation:

We note that the node Q is also between the resistors of 1ohm and 2ohms.

We note that the node P is also between the resistors of 2ohms and 3ohms.

Thus, all these resistors are in parallel, beween nodes P and Q

1/Re=1/R1+1/R2+1/R3

1/Re=1/1+1/2+1/3=(6+3+2)/6=11/6 [ohm^(-1)]

Re=6/11=0.54ohms

Rounding to the tenth: Re=0.5 ohms

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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

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

Using Ohm's Law, the current is:

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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

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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,

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

r=400\Omega is the internal resistance

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

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

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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.

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R=1000 \Omega is the resistance of the resistor

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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\%

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