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2 years ago

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A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can

be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?

1 answer:

Pani-rosa [81]2 years ago

7 0

Answer:a) 4+8+24=36

B) 1/4+1/8+1/24=10

C) yu will connect them in parallel connection.

D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.

Explanation:

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Explanation:

The given data is as follows.

mass, m = 75 g

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First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

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Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

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2 years ago

A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

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Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

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so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

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$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

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2 years ago

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Answer:

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Distance is proportional to focal length, so

d∝f

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$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

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$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

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Answer:

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2 years ago