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garik1379 [7]
3 years ago
8

A 6.00 V battery has an internal resistance of 0.8322 What is the terminal voltage if it is connected in series to a circuit wit

h a total resistance of 7380 O 5.89 V O 591V 5.87V O 5.99
Physics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

The terminal voltage will be 5.99 volt.

(d) is correct option.

Explanation:

Given that,

Voltage = 6.00

Internal r= 0.8322 ohm

Total resistance R =7380 ohm

We need to calculate the current

Using current formula

I=\dfrac{V}{R+r}

Put the value into the formula

I = \dfrac{6}{7380+0.8322}

I=0.000812\ A

We need to calculate the voltage drop due to internal resistance

V' = Ir

V'=0.000812\times0.8322

V'=0.00067\ volt

Now, The terminal voltage will be

V''=6-V'

V''=6-0.00067

V''=5.99\ volt

Hence, The terminal voltage will be 5.99 volt,

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
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Since it is a iso-thermal process therefore q=w

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Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

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Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

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n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

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b) Intrinsic impedances of media 1 and media 2

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n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

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The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

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n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

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You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

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