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monitta
2 years ago
12

A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?

Physics
1 answer:
Naddika [18.5K]2 years ago
6 0

Answer:

0.5 A

Explanation:

Applying,

V = IR.................. Equation 1

Where V = Voltage, I = current, R = Resistance.

make I the subject of the equation

I = V/R............... Equation 2

From the question,

Given: V = 75 V, R = 150 Ω

Substitute these values into equation 2

I = 75/150

I = 0.5 A.

Hence the cuurent through the resistor is 0.5 A

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Answer:

Given:

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Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

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where

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v_{e} = velocity of an electron

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Now,

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\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

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Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

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Now,

Kinetic energy of a proton = thermal kinetic energy

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\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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