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kolbaska11 [484]
3 years ago
10

Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt

Physics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

<em>The drop voltage is 0.3 V</em>

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

\displaystyle i=\frac{\varepsilon }{R_e+R_i}

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

\displaystyle i=\frac{24 }{158+2}

\displaystyle i=\frac{24 }{160}

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

V_i = i.R_i

V_i = 0.15*2

\boxed{V_i = 0.3\ V}

The drop voltage is 0.3 V

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Answer:

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The expression for the amplitude under these conditions is:

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To calculate the angular speed w, we use the following expression:

w = √k/m  (2)

Calculating w:

w = √300/2 = 12.25 rad/s

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b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

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Replacing the data in (6):

Ф = arccos(0/0.98)

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And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

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cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

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Answer:

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Explanation:

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Answer:

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