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2 years ago

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As the resistance increased under 25 V, the current . Compared to calculated currents, experimental currents proved to be . Curr

ents varied proportionally according to resistance: 10 Ω of resistance produced a current of A, while 100 Ω of resistance produced a current of A.

2 answers:

Zanzabum2 years ago

7 0

Answer:

1) 10 Ω of resistance produced a current of 2.5 Ampere.

2) 100 Ω of resistance produced a current of 0.25 Ampere.

Explanation:

Ohm's law is defined as the law which gives us the relationship between voltage and current.

The equation used to represent this law is:

$V=IR$

where,

V = voltage of the circuit. The unit for this is Volts.

I = current of the circuit. The unit for this is Amperes

R = resistance of the circuit. The unit for this is Ohms.

1) 10 Ω of resistance produced a current of:

V = 25 Volts

$V=IR$

$I=\frac{V}{R}=\frac{25 V}{10\Omega }=2.5 Ampere$

2) 100 Ω of resistance produced a current of:

V = 25 Volts

$V=IR$

$I=\frac{V}{R}=\frac{25 V}{100\Omega }=0.25 Ampere$

Rzqust [24]2 years ago

4 0

Answer:

Decreased, The Same, 2.5 and .25

Explanation:

These are the answers

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Answer:

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Explanation:

Here Mass flux of water vapour is given as

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$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

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R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

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Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

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$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

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Answer:

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Explanation:

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Answer:

Explanation:

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Answer:

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Answer:

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