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3 years ago

A milliammeter has an internal resistance of 5ohms and

Physics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

Rs = 0.02008 Ω = 20.08 mΩ

Explanation:

The range of an ammeter can be increased by connecting a small shunt resistance to it in a series combination. This shunt resistance can be calculated by the following formula:

$Rs = \frac{I_gR_g}{I - I_g}$

where,

$R_s$ = value of shunt resistance = ?

$I_g$ = current range of ammeter = 20 mA = 0.02 A

I = Required range of ammeter = 5 A

$R_g$ = Resistance of ammeter = 5 ohms

Therefore,

$R_s = \frac{(0.02\ A)(5\ ohms)}{5\ A-0.02\ A}$

Rs = 0.02008 Ω = 20.08 mΩ

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3 years ago

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Ivanshal [37]

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3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

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The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

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R₂ = infinity

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Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

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Answer:

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